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TWENTIETH   CENTURY  TEXT-BOOKS 


TWENTIETH    CENTURY   TEXT-BOOKS 


THE  YOUNG  AND  JACKSON  ALGEBRAS 

Elementary  Algebra,  adapted  to  one 
and  a  half  or  two  years'  work.  It  com- 
pletely fulfills  college  entrance  require- 
ments  in   the  subject,      xii,  442  pages, 

$1.12. 

A  First  Course  in  Elementary  Alge- 
bra, adapted  to  one  year's  work.  It 
includes  the  maximum  number  of  sub- 
jects contained  in  high-school  first-year 
courses  in  algebra,  ix,  294  pages,  95  cents. 

A  Second  Course  in  Elementary 
Algebra,  designed  for  pupils  who  have 
had  one  year  of  algebra,  and  contains 
no  advanced  algebra,  vii,  206  pages, 
70  cents. 

D.    APPLETON    AND    COMPANY 
NEW  YORK  CHICAGO 


163 


TWENTIETH   CENTURY    TEXT-BOOKS 


A  SECOND  COURSE  IN 


ELEMENTARY    ALGEBRA 


BY 


J.   W.    A.   YOUNG,  Ph.D. 

ASSOCIATE    PROFESSOR    OF    THE    PEDAGOGY    OF    MATHEMATICS 
THE    UNIVERSITY    OF    CHICAGO 


AND 


LAMBERT   L.    JACKSON,  Ph.D. 

FORMERLY    PROFESSOR    OF    MATHEMATICS,    STATE    NORMAL 
SCHOOL,    BROCKPORT,    NEW    YORK 


NEW    YORK 

D.    APPLETON   AND    COMPANY 

1910 


Copyright,  1908,  1910,  by 
D.  APPLETON  AND  COMPANY. 


PREFACE 

The  material  in  this  volume  is  prepared  for  the  use  of  pupils 
who  have  done  a  year's  work  in  Elementary  Algebra.  Many 
high  schools  divide  their  Algebra  work  into  two  courses  sepa- 
rated by  some  work  in  Geometry  and  elementary  science. 
These  schools  often  find  it  more  convenient  and  economical 
to  use  two  books  in  Algebra,  one  for  the  first  course  and  another 
for  the  second.  The  authors'  First  Course  in  Elementary 
Algebra  is  planned  for  first-year  work  and  this  book  is 
planned  for  the  second-year  work. 

The  first  six  chapters  are  a  review  and  extension  of  the 
topics  of  the  First  Course.  The  chapter  on  logarithms  is 
new  in  name  only,  because  in  theory  it  is  an  extension  of  the 
subject  of  exponents.  The  remainder  of  the  volume  treats 
the  usual  topics,  Equations,  Proportion,  Variation  and  Series, 
supplemented  by  problems  applying  Algebra  to  Geometry. 

Throughout  the  treatment  the  authors  have  constantly  kept 
in  mind  both  the  logical  value  and  the  practical  utility  of  the 
subject. 

The  logical  value  of  Algebra  is  of  prime  importance ;  hence, 
the  proofs  of  processes  are  based  upon  reasons,  both  correct 
and  satisfying  to  the  mind  of  the  pupil.  On  the  other  hand, 
subtle  distinctions  and  arguments  savoring  of  higher  mathe- 
matical methods  without  their  true  rigor  have  been  avoided. 

The  utility  of  Algebra  is  given  the  emphasis  which  it  so 
richly  deserves.  This  is  done  by  making  the  equation  promi- 
nent, by  introducing  simple  formulas  of  Geometry  and  Physics, 
and  by  applying  Algebra  to  modern  industrial,  commercial, 
and  scientific  problems  whose  content  can  readily  be  undcr- 

iii 


iv  •  PREFACE 

stood  by  the  pupil.     Useless   puzzles   and  problems  relating 

to  past  conditions  have  been  excluded,  with  the  exception  of 

a  few  supplementary  problems  retained  on  account  of  their 

historical  interest. 

The  Summaries  and  Reviews  at  the  ends  of  chapters  furnish 

systematically,  and  in  small  compass,  the  essentials  of  Algebra. 

By  reference  to  these  the  pupil  can  best  review  and  unify  his 

knowledge  of  the  subject. 

THE   AUTHORS. 


CONTENTS 

CHAP. 

I.  —  Fundamental  Processes     ....         Pp.  1-24 

Definitions,  1  ;  Addition,  1  ;  Graphical  Representation  of  Addition, 
2,  3  ;  Commutative  Law  of  Addition,  2  ;  Associative  Law,'  3  ;  Sub- 
traction, 4 ;  Relative  Numbers,  5 ;  Graphical  Representation  of 
Relative  Numbers,  5  ;  Rules  for  Signs  of  Operation,  6  ;  Absolute 
Value,  6  ;  Addition  of  Relative  Numbers,  7  ;  Subtraction  of  Relative 
Numbers,  8;  Multiplication,  11  ;  Commutative  Law  of  Multiplication, 
11  ;  Factor,  11  ;  Associative  and  Distributive  Laws,  12;  Multiplica- 
tion of  Relative  Numbers,  13  ;  Division,  14  ;  Fractions,  15  ;  Graphical 
Representation  of  Fractions,  15  ;  Properties  of  Fractions,  16  ;  Multi- 
plication of  Fractions,  17  ;  Division  of  Fractions,  19  ;  Complex  Frac- 
tions, 19  ;  Factoring,  20  ;  Review,  23-24. 

II.  —  Equations Pp.  25-41 

Equations  of  One  Unknown,  25-30 ;  Definitions,  25,  26  ;  Equivalent 
Equations,  26  ;  Linear  Form,  26  ;  General  Solution,  27  ;  Equations 
of  Two  Unknowns,  31-33  ;  Systems  of  Equations,  31  ;  Simultaneous 
Equations,  31  ;  General  Solution,  31  ;  Formulas,  32  ;  Equations  of 
Three  or  More  Unknowns,  33-35  ;  Solution,  33  ;  Quadratic  Equa- 
tions, 36  ;  Solution,  36  ;  Review,  39-41. 

III.  —  Kadicals Pp.  42-52 

Definitions,  42 ;  Addition  and  Subtraction,  44  ;  Multiplication  of  Ex- 
pressions containing  Square  Roots,  45  ;  Division  of  Square  Roots, 
46 :  Rationalizing  the  Denominator,  47  ;  Radical  Equations,  48 ; 
Summary,  49;  Review,  50  ;  Supplementary  Work,  51. 

IV.— Exponents Pp.  53-73 

Laws,  53 ;  Fractional  Exponents,  56  ;  Zero  and  Negative  Exponents, 
63  ;  Use  of  Zero,  Negative,  and  Fractional  Exponents,  65  ;  Summary, 
68;  Review,  70;  Supplementary  Work,  71. 

V.  —  Logarithms Pp.  74-89 

Use  of  Exponents  in  Computation,  74  ;  Definitions,  79  ;  Explanation 
of  the  Tables  —  Negative  Characteristics,  80  ;  Use  of  the  Tables,  82  ; 
Summary  —  Review,  88. 

v 


vi  CONTENTS 

CHAP. 

VI.  —  Imaginary  and  Complex  Numbers  .       Pp.  90-99 

Imaginary  Numbei-s  ■ —  Ueal  Numbers,  00  ;  Complex  Numbers,  91  ; 
Addition  and  Subtraction  —  Multiplication,  92  ;  Division,  03  ;  Powers 
of  the  Imaginary  Unit,  04  ;  Imaginaries  as  Roots  of  Equations,  94  ; 
Summary,  96  ;  Supplementary  Work,  97. 

VII. — Quadratic  Equations      ....  Pp.  100-131 

General  Form,  100;  General  Solution  —  Solution  by  Formula,  102; 
Literal  Quadratic  Equations,  103  ;  Collected  Methods,  104  ;  Relation 
of  Roots  to  Coefficients,  107;  Testing  —  Character  of  Roots,  108; 
Discriminant  —  Factoring  Quadratic  Polynomials,  109  ;  Graphical 
Solution,  111  ;  Higher  Equations  solved  by  the  Aid  of  Quadratic 
Equations,  115  ;  Binomial  Equations,  116  ;  Summary,  117  ;  Review, 
118;  Supplementary  Work,  122. 

VIII.  —  Systems    of   Quadratic   and   Higher 

Equations Pp.  132-147 

Simultaneous  Quadratic  Equations,  132  ;  Simultaneous  Higher  Equa- 
tions, 141  ;  Summary,  142  ;  Review,  143  ;  Supplementary  Work, 
145. 

IX.  — Proportion Pp.  148-159 

Definitions  —  Relation  to  the  Equation,  148;  Mean  Proportional,  153; 
Summary,  156  ;  Review,  157  ;  Supplementary  Work,  159. 

X.  —  Variation Pp.  160-169 

Direct  Variation  —  Relation  to  Proportion  —  Expressions  for  Direct 
Variation,  160;  Inverse  Variation  —  Expressions  for  Inverse  Varia- 
tion, 161  ;  Graphical  Work,  163  ;  Summary,  165  ;  Supplementary 
Work,  166. 

XL  —  Series Pp.  170-191 

Series  —  Terms,  170  ;  Arithmetical  Series  :  Common  Difference, 
171  ;  Last  Term  —  Sum  of  an  Arithmetical  Series  —  Formula  for  the 
Sum,  172;  Collected  Results,  172;  Geometric  Series:  Common 
Ratio,  174;  Last  Term  —  Sum  of  a  Geometric  Series  — Formula  for 
the  Sum,  176  ;  Collected  Results,  176  ;  Means  :  Arithmetical  Mean 
—  Geometric  Mean,  178;  Other  Formulas:  Arithmetical  Series, 
179;  Geometric  Series,  182;  Summary  — Review,  184;  Supple- 
mentary Work,  187. 

XII.  —  Zero.     Interpretation  of  Results      .  Pp.  192-198 

Zero  and  its  Properties,  192  ;  Interpretation  of  Results,  197. 
XTIT.  —  Supplement Pp.  199-212 

Geometric  Problems  for  Algebraic  Solution,  199. 


A  SECOND  COURSE  IN 
ELEMENTARY  ALGEBRA 

CHAPTER   I 
FUNDAMENTAL   PROCESSES 

1.  Algebra  is  concerned  with  the  study  of  numbers.  The 
number  of  objects  in  any  set  (for  example,  the  number  of 
books  on  a  shelf)  is  found  by  counting.  Such  numbers  are 
called  whole  numbers  or  integers;  also,  natural  or  absolute 
numbers. 

In  arithmetic,  numbers  are  usually  represented  by  means  of 
the  numerals,  0,  1,  2,  3  .  .  .  9,  according  to  a  system  known  as 
the  decimal  notation,  which  we  take  for  granted  is  here  under- 
stood. In  algebra,  numbers  are  also  represented  by  letters 
either  singly  or  in  combinations. 

2.  Graphical  Representation.  The  natural  integers  may  be 
represented  by  equidistant  points  of  a  straight  line ,  thus  : 


3.  Addition.  If  two  sets  of  objects  are  united  into  a  single 
set  (for  example,  the  books  on  two  shelves  placed  on  a  single 
shelf),  the  number  of  objects  in  the  single  set  is  called  the  sum 
of  the  numbers  of  objects  in  the  two  original  sets.  The  pro- 
cess of  finding  the  sum  is  called  addition.  The  sign,  +,  be- 
tween two  number  symbols  indicates  that  the  numbers  are 
to  be  added.     In  the  simplest  instances  the  sum  is  found  by 


counting. 


2  ELEMENTARY    ALGEBRA 

Thus,  to  find  5  +  7,  we  first  count  5,  and  then  count  7  more  of  the 
number  words  next  following  (six,  seven,  eight,  etc.).  The  number 
word  with  which  we  end  (twelve)  names  the  sum. 

4.  Graphical  Representation.  The  sum  of  two  integers  may 
be  represented  graphically  thus  : 

3  +  5  a  +  b 


i — i — i — i — i — i — : — i — i 

012345678 

Theoretically,  the  sum  of  two  integers  can  in  every  instance  be  found 
by  counting.  But  it  is  not  necessary  or  desirable  to  do  so  when  either 
(or  both)  of  the  numbers  is  larger  than  nine.  In  this  case,  the  properties 
of  the  decimal  notation,  as  learned  in  arithmetic,  enable  us  to  abridge  the 
process  of  counting,  and  to  find  the  sum  very  easily  even  when  the  num- 
bers are  large. 

5.  Commutative  Law  of  Addition.  If  two  sets  of  objects  are 
to  be  united  into  a  single  set,  the  number  of  objects  in  the 
latter  is  obviously  the  same  whether  the  objects  of  the  second 
set  are  united  with  those  of  the  first,  or  those  of  the  first  united 
with  those  of  the  second. 

For  example,  the  number  of  books  is  the  same  whether  those  on  the 
first  shelf  be  placed  on  the  second,  or  those  on  the  second  be  placed  on 
the  first.  This  is  true  because  the  operation  of  transfer  neither  supplies 
nor  removes  any  books. 

In  symbols  :  a  +  b  =  b  -f  a. 

This  fact  is  called  the  commutative  law  of  addition. 

The  letters  a  and  b  are  here  used  to  stand  for  integers,  but  later  they 
will  be  taken  to  stand  for  any  algebraic  numbers,  and  the  law  will  still 
apply. 

6.  Graphical  Representation.  The  commutative  law  may  be 
represented  graphically  thus : 

a  +  i 


FUNDAMENTAL   PROCESSES  3 

7.  Addition  of  Two  or  More  Integers.  If  more  than  two  sets 
of  objects  are  united  into  a  single  set,  the  number  of  objects  in 
the  resulting  set  is  called  the  sum  of  the  number  of  objects  in 
the  original  sets,  and  the  process  of  finding  the  sum  is  called 
addition.  As  in  the  case  of  two  numbers,  the  sum  of  three  or 
more  numbers  may  be  found  by  counting  in  the  simplest 
instances,  and  for  larger  numbers,  the  process  may  be  abridged 
by  use  of  the  properties  of  the  decimal  notation. 

8.  The  commutative  law  likewise  applies  to  the  sum  of  three 
or  more  integers.     That  is  : 

The  sum  is  the  same  for  every  order  of  adding  the  numbers. 


9.  Associative  Law  of  Addition.  If  we  have  three  rows  of 
books,  the  number  of  books  is  the  same  whether  those  in  the 
second  row  are  first  placed  with  the  first  row,  and  then  those  in 
the  third  row  placed  with  these,  or  those  in  the  third  row  placed 
with  the  second,  and  then  all  of  these  with  the  first  row. 

In  symbols:         (a  +  b)  +  c  =  a  +  (&  -f  c). 

This  fact  is  called  the  associative  law  of  addition. 

10.    Graphical  Representation.     The   associative  law  may  be 
represented  graphically  thus: 


a  +  b 


c 


_^_ 


a  C b c  "\ 

V  Y 

a    +    b  +  c 

The  properties  stated  above  are  often  used  to  abridge  calculations. 
Thus,  7  +  4  +  3  +  6,  are  more  easily  added  thus  :  (7  +  3)  +  (4  +  0).  In 
actual  work  the  change  of  order  is  made  merely  by  the  eye. 

ORAL    EXERCISES 
Rearrange  advantageously  and  add: 

1.  8  +  3  +  2  +  7.  4.   ±8x  +  73x  +  2x  +  7x. 

2.  91+43+9.  5.    19y  +  54y  +  6-y  +  y. 

3.  87  +  26  +  13.  6.    73  b  +  186  b  +  14  b. 


4  ELEMENTARY   ALGEBRA 

7.  13  a  +  5  a  +  17  a  +  5  a.  11.  279  £ +  347i +  21  £. 

8.  7  a; +  12  a; +  3  a; +  18  a;.  12.  624 p-f 45p  +  6p  +  52>. 

9.  8//  +  10//  +  7//  +  5?/.  13.  93^  +  9  £  +  7  £  +  £. 

1 0.  23  a  +  6  «  +  2  a  +  4  a,  14.  144  m  +  7  /«,  +  6  m  +  3  m. 

WRITTEN     EXERCISES 

Show  graphically  that : 

1.  11  +4  +  6  =  11  +  (4  +  6). 

2.  8  +  5  =  5  +  8. 

3.  4a  +  56  =  5&  +  4a. 

4.  2  a  +  3  a  +  7  a  =  2  a  +  (3  a  +  7  a). 

11. "  Subtraction.  It  often  happens  that  we  wish  to  know 
how  many  objects  are  left  when  some  of  a  set  are  taken  away, 
or  to  know  how  much  greater  one  number  is  than  another. 
The  process  of  finding  this  number  is  called  subtraction.  The 
number  taken  away  is  called  the  subtrahend,  that  from  which 
it  is  taken,  the  minuend,  and  the  result,  the  difference  or  the 
remainder. 

12.  The  sign  of  subtraction  is  — . 

13.  Subtraction  is  the  reverse  of  addition,  and  from  every 
sum  one  or  more  differences  can  at  once  be  read. 

Thus,  from  5  +  7  =  12  we  read  at  once  12  —  5  =  7, 

12  -  7  =  5. 
And  from  5  +  5  =  10,  we  read  10  —  5  =  5. 

And  from  a  +  b  =  c,  we  read  c  —  a  =  b, 

c  —  6  =  a. 

Likewise,  from  a  +  b  +  c  =  d  we  read  d  —  a  =  b  +  c, 

d  —  (a  +  6)  =  c,  etc. 

14.  There  is  no  commutative  law  of  subtraction.  For  7  —  4 
is  not  the  same  as  4  —  7.  In  fact,  the  latter  indicated  differ- 
ence has  no  meaning  in  arithmetic.  We  cannot  take  a  larger 
number  of  objects  from  a  smaller  number. 


FUNDAMENTAL   PROCESSES  5 

15.  In  algebra,  where  numbers  are  often  represented  by 
letters,  we  may  not  know  whether  the  minuend  is  larger  than 
the  subtrahend  or  not.  For  example,  in  a  —  b,  we  do  not 
know  whether  a  is  larger  than  b  or  not.  But  it  is  desirable 
that  such  expressions  should  have  a  meaning  in  all  cases,  and 
this  is  accomplished  by  the  definition  and  use  of  relative 
numbers. 

16.  The  First  Extension  of  the  Number  System.  Relative 
Numbers.  Whenever  quantities  may  be  measured  in  one  of 
two  opposite  senses  such  that  a  unit  in  one  sense  offsets  a  unit 
in  the  other  sense,  it  is  customary  to  call  one  of  the  senses  the 
positive  sense,  and  the  other  the  negative  sense,  and  numbers 
measuring  changes  in  these  senses  are  called  positive  and 
negative  numbers  respectively.  (For  examples,  see  First 
Course,  pp.  32-34.) 

17.  A  number  to  be  added  is  offset  by  an  equal  number  to  be 
subtracted ;  hence  such  numbers  satisfy  the  above  definition, 
and  numbers  to  be  added  are  called  positive,  and  those  to  be 
subtracted  are  called  negative.  Consequently,  positive  and 
negative  numbers  are  denoted  by  the  signs  +  and  — 
respectively. 

Thus,  +  5  means  positive  five,  and  denotes  five  units  to  be  added  or  to 
be  taken  in  the  positive  sense. 

—  5  means  negative  five,  and  denotes  five  units  to  be  subtracted,  or  to 
be  taken  in  the  negative  sense. 

18.  Graphical  Representation.  Relative  integers  may  be 
represented  graphically  thus: 

-5    -4  -3   -2    -I       0    +1    +2   +3  +4   +5 

It  appears  that  the  positive  integers  are  represented  by  just  the  same 
set  of  points  as  the  natural  or  absolute  integers.  For  this  and  other 
reasons  the  absolute  numbers  are  usually  identified  with  positive  numbers. 
Although  it  is  usually  convenient  to  do  this,  we  have  in  fact  the  three 
classes  of  numbers  :  the  absolute,  the  positive,  and  the  negative.  Thus, 
we  may  consider  $5  without  reference  to  its  relation  to  an  account,  or  we 
can  consider  it  as  §  5  of  assets,  or  we  may  consider  it  as  §  5  of  debts. 


6  ELEMENTARY   ALGEBRA 

19.  The  following  rules  make  clear  in  every  instance 
whether  the  signs  -f,  —  denote  the  operations  of  addition  or 
subtraction,  or  the  positive  or  negative  character  of  the  num- 
bers which  these  signs  precede : 

I.  If  used  where  a  sign  of  operation  is  needed,  the  signs  +,  — , 
shall  be  regarded  as  signs  of  operation. 

For  example  : 

In  8  —  5,  —  is  a  sign  of  operation  (subtraction). 

In  —  8  +  5,  —  is  a  sign  of  character,  because  ho  sign  of  operation  is 
needed  before  the  8,  but  +  is  a  sign  of  operation. 

In  the  problem,  "  Add  —  8  and  +  5,1'  both  the  signs  are  signs  of  char- 
acter, because  no  sign  of  operation  is  needed  ;  the  operation  has  already 
been  named. 

II.  If  it  is  necessary  to  distinguish  a  sign  of  character  from  a 
sign  of  operation,  the  former  is  put  into  a  parenthesis  with  the 
number  it  affects. 

Thus,  —  8  +  (—3),  means  :  negative  8  plus  negative  3. 

III.  When  no  sign  of  character  is  expressed,  the  sign  plus  is 

understood. 

Thus,  5  —  3  means  :  positive  5  minus  positive  3. 
Similarly,  8  a  -f  9  a  means  :  positive  8  a  plus  positive  9  a. 

20.  Absolute  Value.  The  value  of  a  relative  number  apart 
from  its  sign  is  called  its  absolute  value. 

ORAL    EXERCISES 

Read  the  following  in  full,  according  to  the  agreements  of 
Sec.  19 : 

9.  12_(_5). 

10.  -12 -(+5). 

11.  -7- (-9). 

12.  2  a  —  (+3 a). 

13.  c  +  d. 

14.  c  —  d. 

15.  m  +  (—»). 

16.  4ic  +  (—  2x). 


1. 

6-4. 

2. 

-5-8. 

3. 

-  8  +  20. 

4. 

2  a  +  3  a. 

5. 

2b -3b. 

6. 

-2  a -3  a. 

7. 

2y+(+3y). 

8. 

3p-(-2p). 

17. 

7-9. 

18. 

7  +  9. 

19. 

-7  +  (-9). 

20. 

2y-(-3y). 

21. 

-2x-(-3x). 

22. 

-2x  +  (-3x). 

23. 

-2  b  -(-5  c). 

24. 

3a-  (+5 y). 

FUNDAMENTAL    PROCESSES  7 

WRITTEN    EXERCISES 
Indicate,  using  the  signs  +,  —  : 

1.  The  sum  of  positive  5  and  positive '3. 

2.  The  sum  of  positive  a  and  negative  b. 

3  The  difference  of  positive  p  and  positive  q. 

4.  The  difference  of  negative  5  and  positive  3. 

■  5.  The  difference  of  negative  x  and  positive  y. 

6.  The  sum  of  positive  a  and  positive  b. 

7.  The  sum  of  negative  ab  and  negative  ab. 

8.  The  sum  of  positive  y  and  negative  x. 

9.  The  difference  of  positive  xy  and  negative  xy. 

10.  The  difference  of  negative  pq  and  positive  mn. 

21.  Addition  of  Relative  Numbers.  Just  as  3  lb.  +  5  lb.  =  8 
lb.,  so  3  positive  units  4-  5  positive  units  make  8  positive  units, 
and  3  negative  units  4-5  negative  units  make  8  negative  units. 

To  add  units  of  opposite  character,  use  is  made  of  the  de- 
fining property  of  relative  numbers,  that  a  unit  in  one  sense 
offsets  a  unit  in  the  other  sense.  Thus,  to  add  3  positive  units 
and  7  negative  units  we  notice  that  the  3  positive  units  offset 
3  of  the  negative  units  and  the  result  of  adding  the  two  will 
be  4  negative  units. 

That  is, 

(+3)  +  (-7)  =  (4-3)  4-  (-3)+  (-4)  =-4. 
In  general : 

I.  If  two  relative  numbers  have  the  same  sign,  the  absolute 
value  of  the  sum  is  the  sum  of  the  absolute  values  of  the  addends, 
and  the  sign  of  the  sum  is  the  common  sign  of  the  addends. 

11.  If  two  relative  numbers  have  opposite  signs,  the  absolute 
#  value  of  the  sum  is  the  difference  of  the  absolute  values  of  the 

addends,  and  the  sign  of  the  sum  is  the  sign  of  the  addend  having 
the  larger  absolute  value. 


8  ELEMENTARY   ALGEBRA 

22.  More  than  two  numbers  are  added  by  repetition  of  the 
process  just  described.     This  may  be  done  either: 

(1)  by  adding  the  second  number  to  the  Jirst ;  then  the  third 
number  to  the  result,  and  so  on  ;  or 

(2)  by  adding  separately  all  the  positive  numbers  and  all  the 
negative  numbers,  and  then  adding  these  two  results. 

23.  It  may  be  verified  that  the  Commutative  and  the  Associa- 
tive Laws  of  Addition  hold  also  for  relative  integers. 

24.  Subtraction  of  Relative  Numbers.  Since  n  units  of  one 
sense  are  offset  by  adding  n  units  of  the  opposite  sense,  we 
may  subtract  n  units  of  one  sense  by  adding  n  units  of  the 
opposite  sense. 


Thus, 

7-(+3)  =  7  +  (-3). 

And, 

7-(_3)  =  7  +  (+3). 

And, 

4-(+7)  =  4  +  (-7). 

25.  Accordingly  subtraction  may  be  regarded  as  a  variety  of 
addition :   To  subtract  a  monomial,  we  add  its  opjwsite. 

To  subtract  an  algebraic  expression  consisting  of  more  than 
one  term,  we  subtract  the  terms  one  after  another. 

In  general,  to  subtract  any  algebraic  exj)ression  toe  may  change 
the  sign  of  each  of  its  terms  and  add  the  result  to  the  minuend. 

The  subtraction  of  a  larger  number  from  a  smaller  number  (as  in  the 
third  example,  Sec.  24)  is  made  possible  by  the  introduction  of  the  idea 
of  relative  numbers. 

ORAL   EXERCISES 

State  the  sums : 

1.  5 +  (-3).  4.    -12z  +  (-18z). 

2.  -6a+(-7a).  5.    11  x  +  (-  2x)  +  (-5x). 

3.  -lly  +  Zy.  6.     -3q  +  7q  +  (-6q). 

7.    How  may  the  correctness  of  a  result  in  subtraction  be 
tested  ?    State  the  differences  : 


FUNDAMENTAL   PROCESSES  9 

8.  11-6.         10.    -lla-(-Ga).     12.    -31y-(-3y). 

9.  -11-6.    ll.   31a- (+5 a).  13.    17  p  -  (- 17 p). 

14.  How  may  a  parenthesis  preceded  by  the  sign'  +  be  re- 
moved without  changing  the  value  of  the  expression?  One 
preceded  by  the  sign  —  ? 

15.  How  may  terms  be  introduced  in  a  parenthesis  preceded 
by  the  sign  +  without  changing  the  value  of  the  expression  ? 
In  a  parenthesis  preceded  by  the  sign  —  ? 

WRITTEN     EXERCISES 


Acl< 
1. 

1: 
2a  +  5 

6. 

c  +  d—5 

11. 

4  x  —  2  z  +  y 

a +  4 

c  —  a"  4-5 

2  X  —      ?/  +  2 

2. 

3a  +  8 

7. 

x+y+2z 

12. 

1  +  m3  4-  jr 

a  —  4 

x  —  y  -f-  4  z 

1  —  7ii3  —  p2 

3. 

6b  +    c 

8. 

P  +  (/  —       7/1 

13. 

ax  4-  6?/  +  cz2 

3  b -2  c 

^  —  <?  +  2  /// 

bx  -{-ay  —   z2 

4. 

-  3  a .  +     b 

9. 

2  a  —    y  +  z 

14. 

1.5  a;  4-  3.5?/  4-    2. 

2a-3b 

2x  +  2y  —  4z 

.5a4-6.5y—  .lz. 

5. 

4a  —  5 

10. 

ax-\-by  +  c 

15. 

2a'"l""52/—32; 

3a  +  7 

itract : 

aa  4-    #  —  c 

i»-#2/+£« 

Sut 

16. 

4a  +  6 

20. 

7  x  4-  3  ,v 

24. 

41  x2  -  16  f 

2a-9 

2  y  -  4  a; 

15  a2  -  20  y2 

17. 

8  x  4-  3 

21. 

4a  +  3i 

25. 

—  11  m  +  AOp 

-  3  x  +  2 

2c-5a 

-40  m  -12  p 

18. 

llx  —  Ay 

22. 

16  y+       2 

26. 

x  —  7  y  -\-5  z 

19  a 

8  #  - 10  2 

—  x  +  Ay  —  6z 

19. 

12  * 

23. 

10  or9  -  16  y2 

27. 

p  4-     ^  —     m 

-  6  «  +  3 

20  a2  4-    4y2 

6^—  2  g  4- 4  ???, 

10 


ELEMENTARY    ALGEBRA 


28.    ax2  +  by2  +  c 
ax2  —  by2  —  c 


29. 

m2 
—  m2 

+  2p2 
-4p2 

-6g2 
+  5g2 

30. 

40  x2 
50  a;2 

- 10  f 

+  40  7/ 

+  z2 
-lz2 

31.   2.5  a +  6.3  6 -.lc 

1.5  a  —  3.5  b  —  .9  c 


32. 

2"  ^ 
f  35 

33. 

a2a; 
a2a; 

+  &22/  +  A2 

+  by2  +  cz2 

Remove  parentheses  and  unite  terras  as  much  as  possible : 

34.  3  a-26+ (36-7  a). 

35.  4m +  5—  (63  —  3p). 

36.  (lltc  +  52/)-(-3aj  +  22). 

37.  7 -[2 -(3 -5)]. 

38.  (4  a  +  2  o)  -  [(7  a  -  5  a)  +  (-  6  a  - 17  a)]. 

39.  3  +  {5a;-2-(7a;-l-2  +  3a;)}. 

40.  ±x  +  \2x-[3-(7x  +  5)-l+x']\. 

41.  a&-{2a&  +  c-[3c-(6-a&)]|. 

42.  12  xy-\2  xy  -  (3  z  +  [4  av/  -  (2 2  +  a^)]  } • 

43.  _(5M-3^  +  l)-$-2ixy  +  3.^-(aW  +  2)|. 

44.  -  {4  a&c  —  (2  ac  +  6c) ]  +  \ 6  a&c  +  (2  ac  —  6c)}. 

Write  the  expression  of  Exercises  45-56,  as  a;  minus  a 
parenthesis.  Also  group  the  terms  involving  x  and  y  in  each 
expression  in  a  parenthesis  preceded  by  the  sign  — . 


45.  3a  +  x  +  2y. 

46.  —  5y  +  7c  —  8  +  05. 

47.  6«2  +  4?/  +  .x--3. 

48.  2  m  +  p  +  a;  —  7/. 

49.  a  +  x  —  6  +  cy. 

50.  J7  +  X  -  7/ +  &?/. 


51.  x  —  3b+2y—  c. 

52.  3  b  +  as  +  ay  +  by. 

53.  &?/  +  m  —  ?*  +  3J. 

54.  2p  —  q  +  x  —  y. 

55.  a//.  +  .(' —  Z  +  Z?/. 

56.  C  +  fit  +  9)  —  (a  +  6)  ?/. 


FUNDAMENTAL   PROCESSES  11 

26.  Multiplication.  To  multiply  two  absolute  integers  means 
to  use  the  one  (called  the  multiplicand)  as  addend  as  many  times 
as  there  are  units  in  the  other  (called  the  multiplier).  The 
result  is  called  the  product. 

Thus,  3  times  4  means  4  +  4  +  4. 

The  simpler  products  are  obtained  by  actually  making  the  additions  that 
are  implied.  For  large  numbers,  as  we  have  seen  in  arithmetic,  the  pro- 
cess may  be  much  abridged  by  use  of  properties  of  the  decimal  notation. 

27.  Commutative  Law  of   Multiplication.     The  expression   3 
times  5  means  5  +5+5,  and  may  be  indicated  as  follows: 
That  is,  since  each   horizontal    line 

(or  row)  contains  5  dots,  there   are  •     •     •  • 

all   together    3  times  5  dots.      But      (A)      •     •     •     •     • 
each   vertical   line    (or  column)  con- 
tains 3  dots  and  there  are  5  columns. 

Hence  there  are  5  times  3  dots  all  together.  But  the  number  of 
dots  is  the  same  whether  we  count  them  by  rows  or  by  columns, 
hence  5  times  3  equals  3  times  5.     Quite  similarly,  if  we  have 


(B) 


a< 


t       t        t 


a  rows  of  dots  with  b  dots  in  each  row,  it  follows  that  a  times 
b  equals  b  times  a. 

The  result  may  be  stated  in  symbols  thus : 

ab  =  ba. 

This  fact,  called  the  Commutative  Law  of  Multiplication,  means 
that  the  product  is  not  altered  if  multiplier  and  multiplicand 
are  interchanged.  Consequently,  these  names  are  frequently 
replaced  by  the  name  factor  applied  to  each  of  the  numbers 
multiplied. 


12  ELEMENTARY   ALGEBRA 

28.  Let  each  dot  of  the  block  of  dots  of  (A)  above  have  the 
value  of  6.  Since  there  are  3x5  dots,  the  value  of  the  block 
would  be  (3  x  5)  x  6. 

A  second  expression  for  the  value  of  the  block  is  obtained 
by  finding  the  value  of  the  first  row  (viz.  5x6)  and  multiply- 
ing it  by  the  number  of  rows,  or  3.  The  expression  resulting 
must  be  equal  to  that  already  found,  or : 

3x(5  x6)  =  (3x5)x6. 

Similarly,  if  each  dot  of  the  block  (B)  above  has  the  value 

c,  it  follows  that 

a(bc)  =  (ab)  c. 

That  is :  The  product  of  three  absolute  integers  is  not  altered 
if  they  be  grouped  for  multiplication  in  any  way  possible  without 
changing  the  order.  Tliis  is  a  case  of  vohat  is  known  as  the 
Associative  Law  of  Multiplication  for  absolute  integers. 

29.  By  similar  methods  and  use  of  these  results  it  can  be 
proved  that  both  the  Commutative  Law  and  the  Associative 
Law  apply  to  all  products  of  absolute  integers.     That  is : 

Commutative  Law.  The  product  of  any  number  of  given  factors 
is  not  changed,  if  the  order  of  the  factors  be  changed  in  any  ivay. 

Associative  Law.  The  product  of  any  number  of  given  factors 
in  a  given  order  is  not  changed  if  the  factors  be  grouped  in  any 
way. 

The  Distributive  Law.     From  the  block  of  dots  we  see  that 


c\ 


t 

•     •     •- 


t    t    * +        •    r    T    T" 

:     !     i         !         !     : 

4     4     4-- — -4  4**4- 


c(a  +  b)=ca  -\-cb. 

This  is  called  the  Distributive  Law  of  Multiplication,  and  the 
above  proof  covers  the  case  in  which  a,  b,  and  c  are  absolute 
integers. 


FUNDAMENTAL    PROCESSES  13 

30.  Multiplication  of  Relative  Integers.  To  multiply  by  a 
positive  integer  means  to  take  the  multiplicand  as  addend  as 
many  times  as  there  are  units  in  the  multiplier,  and  to  multi- 
ply by  a  negative  integer  means  to  take  the  multiplicand  as 
subtrahend  as  many  times  as  there  are  units  in  the  multiplier. 

Consequently, 

4-3  =  3+3  +  3  +  3  =  12. 
4(-3)=-3  +  (-3)+(-3)  +  (-3)  =  -12. 
(_  4)3  =  _  3  _  3  _  3  -  3  =  - 12. 
(-4)(-3)  =  -(-3)-(-3)-(-3)  =  12. 

So,  generally, 

(+a)(+6)=  +  a&, 

(+a)(-6)=-a6, 

(_a)(+&)=-a&, 

(—  a)(—  &)=  +  a&. 

31.  In  words  :  TJie  product  of  two  (integral)  factors  of  like 
signs  is  positive,  and  of  two  factors  of  unlike  signs  is  negative  ; 
in  each  case  the  absolute  value  of  the  product  is  the  product  of  the 
absolute  values  of  the  factors. 

32.  We  observe  that  the  Commutative  Law  holds  in  this  case 
also. 

For  example  :  (—  b)  a  =  a  ( —  b).  Since  ba  =  ab,  by  the  Commutative 
Law  for  absolute  integers,  and  by  Sec.  30,  (  —  b)a  —  —  ba  —  —  ab  =  a(— b). 

It  may  be  shown  that  the  Associative  and  the  Distributive 
Laws  also  hold. 

ORAL    EXERCISES 

State  the  laws  that  are  applied  in  the  various  steps  of  the 
following  calculations  : 

1.  2-8- 3-5  =  2-5. 8- 3  =  10 -24  =  240. 

2.  5(17  -  2  -  6  c)  =5(17  -  2)  -  5(6  c)  =  (5  -  2)  17  -  (5  -  6)  c 

=  10  •  17  -  30  c  =  170  -  30  c. 

3.  7  6(5  x  +  ab)  =  (7  6)  (5  x)  +  (7  6)  (ab)  =  35  bx  +  7  ab2. 

4.  (a  +  b)  (c  +  d)  =  a(c  +  d)  +  b(c  +  d)  =ac-\-  ad  +  be  +  bd. 


14                             ELEMENTARY  ALGEBRA 

State  the  products : 

7.        Sx  9.    5x2             11.        6xyz 

—  4  xy  2  x2                      —2  x£ 


5. 

ax 

5b 

6. 

-2a 

-3  a2 

13. 

(a  +  b)2. 

14. 

{a  -  c)2. 

15. 

(x  +  yf. 

16. 

(x  —  a)2. 

17. 

(x-2y) 

8.    —  abb              10.    —  afy  12.    —  4  aH 

arc                      —  xy2  —  3  b2t 

18.  (x  +  y)(x-y).      23.  {m  +  2xf. 

19.  (2x--l)2.              24.  (3-2d)(3+2d). 

20.  (2m-f«)2.             25.  (4  -  x)2. 

21.  0-l)(a-  +  l).      26.  -3x(x  +  2).      . 

22.  (l_y)(l  +  y).      27.  (l-2d)(l+2d). 


WRITTEN    EXERCISES 
Multiply  and  test  (see  First  Course,  p. "69)  : 


1. 

2. 

2a  +  3 
2a  +  4 

3  a  -  2  & 
2a-36 

(3  x  -  4  ?/)2. 
(5y-7t)(Sy 
(6  a  +  13  g)2. 

3.  a? +.2 
a-3 

4.  x2-3a-  +  l 

a-2 

(1  + 

(1- 
(&- 

5.  gf-3y4-4 

y-2 

6.  J9-3J  +  *3 
p-2* 

7. 
8. 
9. 

10. 

+  2  0-                11. 
12. 

.»)(2  +  «)(l-.»). 

3)(&4-7)(6-3). 

33.  Division.  Division  is  the  process  of  finding  a  number 
called  the  quotient,  such  that  when  multiplied  by  a  given  num- 
ber, called  the  divisor,  the  product  is  a  given  number,  called 
the  dividend. 

34.  The  fundamental  relation  of  division  is  : 

Divisor  times  Quotient  equals  Dividend. 

From  this  it  follows  at  once  that  if  dividend  and  divisor 
have  the  same  sign,  the  quotient  is  positive,  and  if  they  have 
unlike  signs,  the  quotient  is  negative. 

Division  is  usually  indicated  in  one  of  the  following  forms: 
f  or  8  -r-  2.     Both  are  read :  "  eight  divided  by  two." 


FUNDAMENTAL   PROCESSES  15 

35.  The  Second  Extension  of  the  Number  System;  Fractions. 
We  found  the  natural  or  absolute  integers  by  counting.  We 
defined  the  operation  of  addition  for  these  integers  and  saw 
that  it  was  always  possible.  Next  we  defined  the  operation 
of  subtraction  for  these  integers,  and  found  that  it  was  not 
always  possible.  This  led  us  to  define  relative  numbers  (posi- 
tive and  negative).  In  the  system  of  numbers  as  enlarged  by 
this  first  extension,  we  saw  that  both  addition  and  subtraction 
are  always  possible.  Then  we  defined  multiplication  for  all 
integers  and  saw  that  it  was  always  possible.  We  now  examine 
the  operation  of  division  as  just  defined. 

The  operation  12  -=-  4  is  possible  in  the  system  of  integers 
because  there  exists  an  integer,  3,  whose  product  with  4  is  12. 
But  the  operation  12  ~  5  is  impossible  in  the  system  of  integers, 
since  there  exists  no  integer  whose  product  with  5  is  12. 
This  leads  us  to  define  another  kind  of  number,  the  fraction. 
This  is  done  by  dividing  the  unit  into  b  equal  parts,  and  taking 

a  of  these  parts.    A  symbol  for  the  new  number  is  -,  in  which 

b 

a  and  b  may  be  any  integers.     This  constitutes  our  second 

extension  of  the  number  system. 

CI 

36.  The  new  number,  -,  is  called  a  fraction;  a  is  called  the 

b 

numerator  and  b  the  denominator  of  the  fraction  ;  a  and  b  together 
are  called  the  terms  of  the  fraction. 

37.  Graphical  Representation  of  Fractions.  Fractions  may  be 
represented  by  points  of  the  number  scale  that  we  have  already 
had. 


Thus,  if  the  distance  from  zero  to  1  is  divided  into  b  equal  parts,  and 
then  a  of  these  parts  laid  off  from  zero,  the  end  point  represents  the  frac- 
tion - .     f  In  the  figure  the  fraction  is  - .  )     Fractions  may  also  be  taken 

in  the  negative  sense.  Two  fractions  are  said  to  be  equal,  or  to  have  the 
same  value,  when  they  are  represented  by  the  same  point  of  the  number 
scale. 


16  ELEMENTARY   ALGEBRA 

38.  If  each  of  the  b  equal  parts  is  halved,  making  2  b  parts, 
and  each  of  the  a  parts  taken  is  halved,  making  2  a  parts,  the 
end  point  remains  the  same. 

That  is : 

a_2o 

b~2b' 
Similarly : 

a  _  na 

b      nb 

(A)  In  words :  TJie  value  of  a  fraction  is  not  altered  if  both 
numerator  and  denominator  is  multiplied  by  the  same  integer. 

(B)  Reading  the  above  equation  from  right  to  left.  The 
value  of  a  fraction  is  not  altered  if  both  numerator  and  denomi- 
nator be  divided  by  a  common  integral  factor. 

Every  integer  may  be  regarded  as  a  fraction. 
Thus,  3  =  f  or  §  or  -1/,  etc. 

39.  These  properties  enable  us  to  add  and  subtract 
fractions. 

If  the  given  fractions  have  the  same  denominator,  we  say  8  fifths  plus 
3  fifths  are  11  fifths  (or  |  +  f  =  Y)  just  as  we  say  8  ft.  +  3  ft.  =  11  ft. 

If  they  have  not  the  same  denominator,  they  are  first  reduced  to  the 
same  denominator,  and  then  the  results  added. 

40.  To  multiply  a  fraction  by  an  integer  we  extend  the 
definition  of  multiplication  to  this  case,  and  we  have : 

3(f)  =  t  +  f  +  f  =  ¥- 

c(cS)  =  ° +  «+...  to  (c  terms)  =«  + "  +  «  "  to  (ca's^ca. 
\bj     b      b  K  }  b  b 


(-cHf) ==_  I  ~  i '" to  ^c  terms^ 


—  a  —  a  —  a'-'  to  (c  terms)  _    _ca 
~b~  "     ~b" 


FUNDAMENTAL   PROCESSES  17 

41.  That  is,  to  multiply  a  fraction  by  an  integer  we  multiply 
the  numerator  by  that  integer. 

The  fraction  -  is  the  quotient  called  for  in  the  indicated  division  a  -f-  b, 

because  the  product  of  the  divisor  b  and  the  asserted  quotient  -  is  the 
dividend  a,  as  seen  in 

h  .  «  _  fa*  _  a.  (by  property  B,  Sec.  38). 
b       b 

In  our  present  number  system  the  fraction  -  may  therefore  always  be 

b 

regarded  as  indicating  the  division  of  a  by  &  (provided  b  is  not  zero). 

42.  The  operation  of  division  is  thus  seen  to  be  possible  in 
the  new  number  system  for  any  integral  dividend  and  any 
integral  divisor  (except  zero). 

We  shall  see,  under  Division  of  Fractions,  that  if  either  dividend  or 
divisor  or  both  are  fractions,  the  operation  is  still  possible  without  en- 
larging the  number  system  further. 

43.  Multiplication  of  Fractions.  The  product  of  two  or  more 
fractions  is  a  fraction  whose  numerator  is  the  product  of  the  given 
numerators  and  whose  denominator  is  the  product  of  the  given 
denominators. 

44.  Since  an  integer  or  an  integral  expression  may  be  re- 
garded as  a  fraction  with  denominator  1,  this  definition  applies 
also  when  one  or  more  of  the  factors  are  integral. 

45.  The  Associative,  Commutative,  and  Distributive  Laws  of 
Multiplication  may  be  shown  to  apply  to  fractions  as  well  as 
to  integers. 

WRITTEN    EXERCISES 

Perform  the  indicated  operations,  and  express  the  result  in 
lowest  terms : 

■   "+1 


5     10  a 

4<x_5a  l 

'    ~5       T*  ""    x  +  1      2(3  +  1) 


o 

o 

o 

6-1 

6  +  1 

x-1 

x  +  1 

18  ELEMENTARY   ALGEBRA 


5.  *L  +  <ty.  7.      2  3P 

2y     4x  a+p     (a+P) 

6.  -$—  +  -•  8.    -  +  — , 

a  —  bb  x*     2xr 

9.     J^.+     5 


,2 


1-g2     l  +  g 

io. %-: +      ''   1 -  + 


(l_SB)(l-y)       (y-a^-l)  ">-!)(!  ^-y) 


12. 


a2—  (6  +  c)a+&c      cr—(b-\-d)a-\-bd     a'2—(c  +  d)a+cd 
2  a1?/  +  y2 '_      y         x2  4-  5  x?/ 


,  _       5        14  .        &      ac4 

13.  —  • 17.     oc  •  -    •  ■ 

7a     156  c2     2&3 

x3     3  xy  ,.      -3a      41 

14.  —    •    — —  •  lo.     •   • 

y"      4Z5  a^-9      9l 

16.  *•  ._•£!.  ijs..  20.  ..<i-lN 

3?/      16ar3y  15i3a;  \        x 


21.   pgr 

22 


Kf-*J- 


a(l  —  a)  1  -f-a 

l+2a  +  a2  '  l-2a  +  a2' 

1-a2  1  -  &2        /a &_\ 

1  +  2  6  +  62  '  62  -  2  a&  +  a2  \JL  -  a      1  -  &/ 
t  + 1      *  _  1        4«2\^  +  2i  +  l 


23. 

24 

'*-l      ?  +  l      1-^V         4< 

tf  +  £-l_£  +  lY!!-.£ 

^r      v2     p     v        J\'p      v 


46.    Reciprocals.     If  the  product  of  two  numbers  is  1,  each 
is  called  the  reciprocal  of  the  other. 


«  ._  b    _. a      fe     «&  _  j 


Thus,  the  reciprocal  of  -  is  - ,  since  —  •  -  =  — 

b       a  b      a     ab 


FUNDAMENTAL   PROCESSES  19 


47.    Division  of  Fractions.     To  divide  -  by  -   means  to  find 

b         a 

a  number  q  such  that  -  ■  q  =  —  • 

a  b 

(1  c 

Multiplying  both  members  by  -,  the  reciprocal  of   -,  we 

,  ,   .  cl     a      ad 

obtain  Q  =  -  '  v  =  v~ ' 

c      b       be 

That  is :  To  obtain  the  quotient,  multiply  the  dividend  by  the 
reciprocal  of  the  divisor. 

We  have  thus  seen  that  in  the  number  system  as  now  extended  the 
four  fundamental  operations  are  possible  when  any  or  all  of  the  numbers 
involved  are  fractions.     (Division  by  zero  is  always  excepted.) 


48.  Complex  Fractions.  If  one  or  both  of  the  terms  of  a  frac- 
tion are  themselves  fractions,  the  given  fraction  is  called  a 
complex  fraction.  Since  a  fraction  indicates  division,  a  com- 
plex fraction  may  be  simplified  by  performing  the  indicated 
division  of  its  numerator  by  its  denominator. 


WRITTEN    EXERCISES 
Divide,  and  express  results  in  lowest  terms : 

5.  ^2c2.  9.  £+f£. 

46  2        8 


7h  2  m  v° 


3x  n     -6  .    -18 


1. 

2  "  6 

2. 

ab 

Ab2 

18  c  ' 

Sac 

3. 

ax2 
2tf  ' 

5x 

bx3^ 
ay2 

4. 

3tf 

2x 

4t/  +  l      2x— 1  Xs         x7 


x  5  a 

8-  r  12- »y 

9  m3  x  2  6 


20  ELEMENTARY    ALGEBRA 

4rc  1  f 

13.    Tjf  14.    f  15.    l" 

16.  f^.aV/i_r 
«+j^(i_™ 


18. 


1      1 


cr/     \tr      a3 


Perform  the  operations  indicated : 

19.  *_  J_.      20.    t+J—      21.    f£=2+lV/ri---2£-Y 

1+1  1+1  \P+<!       J     \       P+9J 

x  t 

49.  Factoring.  The  process  of  finding  two  or  more  factors 
whose  product  is  a  given  algebraic  expression,  is  called  factoring 
the  given  expression. 

In  division  one  factor  (the  divisor)  is  given  and  another  factor  (the 
quotient)  is  to  he  found  such  that  the  product  of  the  two  factors  is  the 
given  expression  (the  dividend).  Division  is  thus  really  a  variety  of 
factoring,  although  the  name  "factoring"  is  not  usually  applied  to  it. 
As  a  rule,  when  an  expression  is  to  be  factored,  none  of  the  factors  is 
specified  in  advance,  and  any  set  of  factors  is  acceptable  on  the  sole  con- 
dition that  their  product  is  the  given  expression. 

50.  Several  type  products  are  of  special  importance.  These 
have  been  treated  in  the  First  Course  (pp.  76-95,  175-188), 
and  are  collected  here  for  reference. 

I.  xy  +  xz  =  x(y  +  z). 

II.  x2±2xy  +  f  =  (x±$f:    >_+  ; 

III.  x>-tf=(x  +  y)(x-y). 

IV.  x2  +  (a  +  b)x  +  ab  =  (x  +  a) (x  +  &). 
V.  Xs  +  3  x-y  +  3  xy°-  +  f  =  (x  +  y)3. 

VI.  or5  +  ?/3  =  (a  +  y)  (x2  —  xy  +  y2). 

VII.  x>-y*  =  tx-y)tx2  +  xy  +  tf). 


FUNDAMENTAL    PROCESSES  21 

In  all  of  these  formulas  the  letters  may,  of  course,  represent  either 
positive  or  negative  numbers. 

Detailed  treatment  of  these  types  is  given  in  the  First  Course,  together 
with  exercises  under  each  type.  The  following  set  of  miscellaneous  exer- 
cises covers  all  the  types. 

ORAL  EXERCISES 
Factor : 

1.  ax  +  ay.  26.  1—2  (x  +  ?/)  +  (x  +  y)2. 

2.  ax  +  a.  27.  49  +  14  x  +  x2. 

3.  ax  +  a2.  28.  9  -  12  y2  +  4  y\ 

4.  abx  —  ay.  29.  16  -  40  z  +  25  z2. 

5.  x2  +  ax.  30.  a2b2  +  2  abed  +  c2d2. 

6.  x^  +  ax2.  31.  \  +  t  +  t2. 

7.  a2J?  —  ax.  32.  ?/ +  .4  /  +  .04. 

8.  mx  +  my  +  m-  33.  m2a?2  +  4  mx  +  4. 

9.  jxv2  +  jrx -{- }ig. 

10.  a8  —  ar  +  x. 

11.  x2y2  +  xy  +  y. 

12.  (a-f-6)ic—  (a +  &)?/. 

13.  s2£  +  ^2  +  s2*2. 

14.  ±gt2  +  gt. 

15.  a&c  —  acd  +  bed. 

16.  a2x2-b2y2. 

17.  4a:2-9z2. 

18.  atbW-c2. 

19.  l-p2q2y2. 

20.  m2p2-sH2. 

21.  («  +  &)2  -  (c  +  tf)2. 

22.  1  -  (m  +  p)\ 

23.  a2  +  2  abc  +  62c2. 

24.  a4.*;4  +  2  a2<B?  +  1. 

25.  a2  +  2  a  (6  +  c)  +  (6  +  c)2. 


34. 

16     2 

35. 

x2  —  5  x  +  6. 

36. 

ar  +  7a;  +  12. 

37. 

:c2  -  x  -  12. 

38. 

a2  +  x  —  6. 

39. 

«2  -  12  x  +  35. 

40. 

s2  _  2  x  _  35. 

41. 

12  a2  +  7  a?  +  1. 

42. 

15y2-2x-l. 

43. 

6  a,-2  -12a;  +  6. 

44. 

3?/2  +  8?/  +  5. 

45. 

a3  —  &3. 

46. 

a3  +  b\ 

47. 

8  -  a363. 

48. 

64  —  a;3?/3. 

49. 

8  a?  -  12  a;2  +  6  x  -  1. 

50. 

27ys  +  27tf  +  9y  +  l 

22  ELEMENTARY   ALGEBRA 


WRITTEN     EXERCISES 
Factor : 

1.  x2  —  49.  26.  my  +  4  s4. 

2.  t2-7t  +  6.  27--  —  +  — -1 

3.  of-3tf-  +  y.  "  8~     4         2 

4.  4v2-9?*2.  28.  .125  r3  +  .75  z2  + 1.5  a,-  +  1. 

5.  1  +  10  a +  25  a2.  29.  (2  m  +  If  -(3  m  -  l)2. 

6.  a3  +  3a2&  +  3«52+63.  30.  8a3  +  12a2£  +  6a£2  +  £3. 

7.  (x  +  l)2+2(a;  +  l)  +  l.  31.  ^  -  16. 

8     1_?  +  I.  32"  f'7/2-Hoz/  +  24. 

a      a2  33.  4  /r  -  12  lit  +  9£2. 

9.    ^-2  a -15.  34.  1-m3. 

10.  a2&2-l.  35.  (2a  +  6)2-(3a-2&)2. 

11.  x4  +  6x2y  +  9y2.  36.  8  a3 -32  a5. 

12.  vs -t\  37  10  m2.r  -  60  mx  +  90  x. 


13.  1  —  »y. 

14.  a4_4a^2. 


38.  ^3  +  27. 

39.  a4 +  3  a2 -28. 


15.  (x  +  y)4  —  1. 

16.  „*-*  40-   8»s-60^  +  150W 


17.  »4  — 2«y  +  ?/4. 

18.  (m+p)3—  1. 

19.  (a;  +  7/)2-(i>  +  ^ 

20.  (a  +  ^/'  +  iB3. 


-125w3. 

41.  .s3-64. 

42.  56  a2  -  68  a  +  20. 

43.  24  a^  +  av- 10. 


2i.   ajs  +  fa^  +  fa  +  f  44-    15  ^  -  34  as- 16. 


22.  a3-lia2  +  fa-i  45-  aV-7cK!  +  12. 

23.  cc2  +  (a  +  &)x-  +  a6.  46.  y4  -  13  ?/2  +  42. 

24.  y2  +  {ac-bd)y-abcd.  47.  (a  + 1)2  -  (a  -  l)2. 

25.  acx2  +  (c6  +  ad)x  +  6(i.  48.  a2-^-}. 

Calculate : 

49.   272-252.  50.   3872-3772.  51.   263-253. 

Note.     Chapters  I  and  II  ai'e  themselves  in  the  nature  of  summaries  ; 

consequently,  no  summaries  of  these  chapters  are  given  here. 


^\DAMENTAL   PROCESSES  23 

REVIEW 

WRITTEN   EXERCISES 

Perform  the  indicated  operations,  expressing  fractional  re- 
sults in  lowest  terms: 

1.  2a -36 4- 4 a  +  11  c  —  2d  +  6-8a-9&  +  3 c  —  4  —  5 d 

4-2a-&. 

2.  afy  +  3  a:?/2  +  4  a,-2  -  2a#  -  3  x2y  +  2  x-y2  -4xy  +  8  xy2 

—  7y2  —  3  x2y2. 

3.  5 x - ly  —  (3a;  +  4y- 2)  +  3  +(8oj - 7)- {2x  —  Sy 

+ 13)4-8(2  a;-l). 

4.  4<c-{2c-(3a;4-2#  +  5c)}. 

5.  5ac-[4&  +  2c[3a-5&-(6a-2&- 7c +  3)] } . 
If  Z  =  4a;  +  2y-3,    M=x-9y  +  l,    R  =  y-5x,find: 


6.  Z  +  Jf+#.  8.  LM-3E. 

7.  3£-2Jf+#.      9.    V--M2. 


10.  (Z  +  7lf)2. 

11.  3L2-5ME. 


If  X  =  7a4-2&,    F=2a-96  +  3,    Z  =  4  & -7a-3,  find: 
12.    X-(3F-2Z).  13.   2F-[Z+3(4X-8F)]. 

i4.  z2-i6F2.      15.  r2  +  4  rz.         i6.  xra. 

,  _    a  -f-  5      a2  +  75  ,  a  —  5 

17. -\ • 

a  —  5      ctr  —  25      a  +  5 

6  m  4-  4  /?  _  3  w2  +  4  jj2      ra  —  2p 
157»  15  mp 

9  -f-  a      a  —  3 


18. 

19.    1 
21. 


1  +  a2      a  + 1 
2  .  1 


5_p 
20. 


H)(-+* 


+w 


p  +  9t  +  20     t2  +  7t  + 12      «■  + 11 1  +  28 

1    /Ca8  +  9a: 


22.    Vaf_3a;  +  14 
Vc3       c2        c 


23. 


3a2V    4a +  2 


24.  (&  +  3)3-(&-2)3. 

25.  (xn  +  3a;"-1  +  5a;n-2)(ar2  +  4a;). 
10  a,-3      5  x3      5  aA     15  x 


26. 


3  ?n2      9  m2 


w 


war 


24  ELEMENTARY   ALGEBRA 

27. 
28. 


33. 


34. 


35  +  12 

5y2  ,  . 

'£  +  ^2 
»  +  02 

.  56  +  1 

xj 

5g  +  g2 

15  +  8 
1      3 

"  2±  +  llg  +  g* 

««2  ™    g— 1     o+l 

29-   o b'  30-   ^m — ^-r' 

2      3  g+1     g— 1 

a8     a  q  —  lq  +  1 

Q1      /3  a      llo\/3a  6\     /3  a     2  6\2 


32. 


6        3  a  J  \  b        3  a)     \b       3  a 
2  13 


<e2  +  4»  +  3      a;2  +  3z  +  2      a2  +  5x-  +  6 

or?/         2  a*3 
a3  +  1  '  a  +  1 

4a  +  7         .  16a2  +  56a  +  49 
a2-16x  +  64  '  a2-64 


35.    -i^+ § +  ^- 


CHAPTER  II 

EQUATIONS 

EQUATIONS   WITH  ONE   UNKNOWN 

51.  Two  algebraic  expressions  are  equal  when  they  represent 
the  same  number. 

52.  If  two  numbers  are  equal,  the  numbers  are  equal  which 
result  from : 

1.  Adding  the  same  number  to  each. 

2.  Multiplying  each  by  the  same  number. 

Subtraction  and  division  are  here  included  as  varieties  of  addition  and 
multiplication. 

53.  The  equality  of  two  expressions  is  indicated  by  the 

symbol,  =,  called  "the  sign  of  equality." 

54.  Two  equal  expressions  connected  by  the  sign  of  equality 
form  an  equation. 

55.  Such  values  of  the  letters  as  make  two  expressions  equal 
are  said  to  satisfy  the  equation  between  these  expressions. 

56.  Equations  that  are  satisfied  by  any  set  of  values  what- 
soever for  the  letters  involved  are  called  identities. 

57.  Equations  that  are  satisfied  by  particular  values  only 
are  called  conditional  equations,  or,  when  there  is  no  danger  of 
confusion,  simply  equations. 

58.  The  numbers  that  satisfy  an  equation  are  called  the 
roots  of  the  equation. 

3  25 


26  ELEMENTARY   ALGEBRA 

59.  To  solve  an  equation  is  to  find  its  roots. 

60.  The  letters  whose  values  are  regarded  as  unknown  are 
called  the  unknowns. 

61.  The  degree  of  an  equation  is  stated  with  respect  to  its 
unknowns.  It  is  the  highest  degree  to  which  the  unknowns 
occur  in  any  terra  in  the  equation.  Unless  otherwise  stated, 
all  the  unknowns  are  considered. 

62.  An  equation  of  the  first  degree  is  called  a  linear  equation. 

63.  An  equation  of  the  second  degree  is  called  a  quadratic 
equation. 

64.  An  equation  of  the  third  or  higher  degree  is  called  a 
higher  equation. 

65.  In  order  to  state  the  degree  of  an  equation  its  terms 
must  be  united  as  much  as  possible. 

66.  Terms  not  involving  the  unknowns  are  called  absolute 
terms. 

67.  Equivalent  Equations'.  If  two  equations  have  the  same 
roots,  the  equations  are  said  to  be  equivalent.  If  two  equa- 
tions have  together  the  same  roots  as  a  third  equation,  the  two 
equations  together  are  said  to  be  equivalent  to  the  third. 

68.  The  Linear  Form,  ax  +  b.  Every  polynomial  of  the  first 
degree  can  be  put  into  the  form  ax  + 1>.  That  is,  by  rearrang- 
ing the  terms  suitably,  it  can  be  written  as  the  product  of  x  by 
a  number  not  involving  x,  plus  an  absolute  term.  Hence,  the 
form  ax  +  6  is  called  a  general  form  for  all  polynomials  of  the 
fir.st  degree  in  x. 

69.  Every  equation  of  the  first  degree  in  one  unknown  can 

be  put  into  the  form  : 

ax  +  6  =  0. 

Consequently  this  is  called  a  general  equation  of  the  first  degree 
in  one  unknown. 


EQUATIONS  27 

i 

70.    General  Solution.     From  the  equation  ax  -\-b  =  0,  (1) 

we  have  ax  =  —  b,  (2) 

and  hence,  x  = (3) 

a 

71. is  the  general  form  of  the  root  of  the  equation  of 

a  '    ■ 

the  first  degree.     There  is  always  one  root,  and  only  one. 

The  advantage  of  a  general  solution  like  this  is  that  it  leads  to  a  for- 
mula which  is  applicable  to  all  equations  of  the  given  form. 

In  words :  When  the  equation  has  been  put  into  the  form 
ax -\-b  =  0  the  root  is  the  negative  of  the  absolute  term  divided  by 
the  coefficient  of  x. 

Test.  The  correctness  of  a  root  is  tested  by  substituting  it 
in  the  original  equation. 

If  substituted  in  any  later  equation,  the  work  leading  to  the  equation 
is,  of  course,  not  covered  by  the  test. 

Results  for  problems  expressed  in  words  should  be  tested 
by  substitution  in  the  conditions  of  the  problem. 

If  tested  by  substitution  in  the  equation  only,  the  correctness  of  the 
solution  is  tested,  but  the  setting  up  of  the  equation  is  not  tested.  Nega- 
tive results  that  may  occur  in  such  problems  are  always  correct  as  solu- 
tions of  the  equations,  but  they  are  admissible  as  results  in  the  concrete 
problem  only  when  the  unknown  quantity  is  such  that  a  unit  of  the  un- 
known quantity  is  offset  by  a  unit  of  its  opposite. 

For  example,  if  the  unknown  measures  distance  forward,  a  negative 
result  means  that  a  corresponding  distance  backward  satisfies  the  con- 
ditions of  the  problem.  But,  if  the  unknown  is  a  number  of  men,  a 
negative  result  is  inadmissible,  since  no  opposite  interpretation  is  possible. 


ORAL  EXERCISES 
Solve  for  x : 

1.  3x  =  15.  4.    .r--6  =  10 

2.  2  a;  =  11.  5.   .r  +  6  =  12. 

3.  4Jx  =  9.  6.    2  £  +  1  =  13. 


28  ELEMENTARY   ALGEBRA 

Solve  for  t : 

7.  6*  =  36.  10.  3t-8  =  22. 

8.  t  —  5  =  20.  11.  at  =  ab. 

9.  2  t  +  5  =  25.  12.  a«  +  6  =  c. 

Solve  for  y : 

13.  3y-l  =  2.  16.  by  =  bc. 

14.  2y  —  l  =  7.  17.  %  =  6  +  c. 

15.  5^  +  5  =  35.  18.  ay-b  =  c. 

Solve  for  a : 

19.  6  a  =  18.  22.  3a  +  l  =  13. 

20.  2|a  =  10.  23.  2  a- 5  =  15. 

21.  a  +  l=13.  24.  5  a  —  6  =  <?. 

WRITTEN    EXERCISES 

Solve  for  x : 

1.  x- 75  =  136.  11.  325  a; +  60  =  400. 

2.  2  a?  -15  =  45.  12.  125  a; +  5  =  10. 

3.  3.r+6  =  48.  13.  8 a? +  625  =  105. 

4.  5  a;  —  10  =  55.  14.  ax  +  6.c  =  c. 

5.  1.5  a;  —  5  =  70.  15.  abx  +  cwc  =  a&. 

6.  3.5  x  —  5  =  100.  16.  c;r  +  fte  =  c  +  d. 

7.  2.1  as  —  41  =  400.  17.  mx  +  px  =  p  +  q. 

8.  1.3  as  +  .1  =  1.79.  18.  lx  +  mx  =  l  —  m. 

9.  .25  a;  +.50  =3.25.  19.  ax  +  bx  =  2(a  +  b). 
10.   .11  x  +  .11  =  1.32.  20.  2cx  +  dx  =  l. 

Solve  and  test : 

21.    2(*-l)+3(2a;  +  5)=0. 


EQUATIONS  29 

22.  4fe±a  +  8_8.  +  t  24.   »1±1_1=|1. 

5  4  2/  —  2      4  —  4  ?/ 

23.  ^l  +  x^l  +  ^zl=x.  25.    l+3psa17-5p> 

2  3  4  P  P 

Solve  for  c : 

26.    5c  +  3a  =  aC.  2g    v==ct  +  9*. 


27. 


30. 


4  c  - 1      2  c  +  5 


L' 


3  m  6  m  29.    (a  +  c)(a  —  c)  =—  (c  +  a)2- 

3C        c-2  =  (c-2)(2c-7).  ^     =z 

c-3      c-4        c*-7c  +  12  '    1  +  ct 


32.  An  inheritance  of  $  2000  is  to  be  divided  between  two 
heirs,  A  and  B,  so  that  B  receives  $  100  less  than  twice  what 
A  receives.     How  much  does  each  receive  ? 

33.  To  build  a  certain  staircase  20  steps  of  a  given  height 
are  required.  If  the  steps  are  made  2  inches  higher,  16  steps 
are  required.     Find  the  height  of  the  staircase. 

34.  In  a  certain  hotel  the  large  dining  room  seats  three 
times  as  many  persons  as  the  small  dining  room.  When  the 
large  dining  room  is  f  full  and  the  small  dining  room  ^  full, 
there  are  100  persons  in  both  together.  How  many  does  each 
room  seat? 

35.  Tickets  of  admission  to  a  certain  lecture  are  sold  at  two 
prices,  one  25  cents  more  than  the  other.  When  100  tickets 
at  the  lower  price  and  60  at  the  higher  price  are  sold,  the  total 
receipts  are  $  95.     Find  the  two  prices. 

36.  Originally,  \^-  of  the  area  of  Alabama  was  forest  land. 
One  third  of  this  land  has  been  cleared,  and  now  20  million 
acres  are  forest  land.  Find  the  area  of  Alabama  in  million 
acres. 

37.  In  a  recent  year  the  railroads  of  the  United  States 
owned  70,000  cattle  cars.  Some  of  these  were,  single-decked, 
and  others  double-decked.  There  were  44,000  more  of  the 
former  than  of  the  latter.    Find  how  many  there  were  of  each. 


30  ELEMENTARY    ALGEBRA 

38.  The  average  number  of  sheep  carried  per  deck  is  45 
larger  thah  the  average  number  of  calves.  If  a  double-decked 
car  has  the  average  number  of  calves  on  the  lower  deck  and  of 
sheep  on  the  upper  deck,  it  contains  195  animals.  Find  the 
number  of  sheep  and  of  calves. 

39.  The  average  number  of  inhabitants  per  square  mile  for 
Indiana  is  \  of  that  for  Iowa,  and  that  for  Ohio  is  32  greater 
than  that  for 'Indiana,  and  62  greater  than  that  for  Iowa. 
Find  the  number  for  each  state. 

40.  Lead  weighs  |4  times  as  much  as  an  equal  volume  of 
aluminium.  A  certain  statuette  of  aluminium  stands  on  a  base 
of  lead.  The  volume  of  the  base  is  twice  that  of  the  statuette, 
and  the  whole  weighs  282  oz.  Find  the  weight  of  the  statuette 
and  of  the  base. 

41.  A  man  inherits  $10,000.  He  invests  some  of  it  in 
bonds  bearing  3^  °J0  interest,  the  rest  in  mortgages  bear- 
ing 51  %  interest  per  annum.  His  entire  annual  income 
from  these  investments  is  $  510.  Find  the  amount  of  each 
investment. 

42.  A  pile  of  boards  consists  of  inch  boards  and  half-inch 
boards.  There  are  80  boards  and  the  pile  is  58  in.  high. 
How  many  boards  of  each  thickness  are  there  ? 

43.  How  much  water  must  be  added  to  30  oz.  of  a  G  °J0  solu- 
tion of  borax  to  make  a  4  %  solution  ? 

44.  How  much  acid  must  be  added  to  10  quarts  of  a  2  °J0 
solution  to  make  a  5%  solution  ? 

45.  A  hardware  dealer  sold  a  furnace  for  $180  at  a  gain  of 
5  %.     What  did  the  furnace  cost  him. 

46.  A  merchant  sold  a  damaged  carpet  for  $42.50  at  a  loss 
of  15  °Jo-     What  did  the  carpet  cost  him  ? 

47.  A  collector  remitted  $475  after  deducting  a  fee  of  5%. 
How  many  dollars  did  he  collect  ? 

48.  The  amount  of  a  certain  principal  at  4  %  simple  interest 
for  1  year  was  $410.     AVhat  was  the  principal? 


EQUATIONS  31 

EQUATIONS  WITH  TWO  UNKNOWNS 

72.  Systems  of  Equations.  Two  or  more  equations  con- 
sidered together  are  called  a  system  of  equations. 

73.  Simultaneous  Equations.  Two  or  more  equations  are  said 
to  be  simultaneous  when  all  of  them  are  satisfied 'by  the  same 
values  of  the  unknowns. 

74.  All  systems. of  two  independent  simultaneous  equations 
of  the  first  degree  in  two  unknowns  can  be  solved  by  the 
method  of  addition  and  subtraction,  which  consists  in  multiplying 
one  or  both  of  the  given  equations  by  such  numbers  that  the 
coefficients  of  one  of  the  unknowns  become  equal.  Then  by 
subtraction  this  unknown  is  eliminated,  and  the  solution  is 
reduced  to  that  of  a  single  equation. 

If  the  coefficients  of  one  unknown  are  made  numerically 
equal,  but  have  opposite  signs,  the  equations  should  be  added. 

75.  Occasionally  the  method  of  substitution  is  useful.  •  This 
consists  in  expressing  one  unknown  in  terms  of  the  other  by 
means  of  one  equation  and  substituting  this  value  in  the  other 
equation,  thus  eliminating  one  of  the  unknowns. 

This  may  be  the  shorter  method  when  an  unknown  in  either 
equation  has  the  coefficient  0,  + 1,  or  —  1. 

Examples  of  the  use  of  these  methods  may  be  found  in  the 
First  Course,  pp.  117,  119. 

General  Solution.  A  general  form  for  two  equations  of  tlie 
first  degree  is 

ax  -\-by  =  e,  (1) 

cx  +  dy=f.  (2) 

From  these  it  is  possible  (without  knowing  the  values  of  a, 
b,  c,  d,  e,  /)  to  find  a  general  form  for  the  solution,  namely : 

x=de-V1  (3) 

ad  —  be 

J     ad  -be  w 


32 


ELEMENTARY   ALGEBRA 


These  results  are  the  formulas  for  the  roots  of  any  system 
of  two  independent  linear  simultaneous  equations  with  two 
unknowns. 

76.  The  application  of  these  formulas  is  made  easier  by 
noticing  how  they  are  formed  from  the  known  numbers  in  the 
equations. 

1 .  The  denominator  is  the  same  in  each  result  and  is  made 
up  from  the  coefficients  as  follows : 


Coefficients  of  x 
a 


Coefficients  of  y 
b 


2.   The  numerator  of  the  value  of  x  is  made  up  thus 


Absolute  Terms 
e 


Coefficients  of  y 
b 


3.   The  numerator  of  the  value  of  y : 


Coefficients  of  x 
a 


Absolute  Terms 
e 


Examples  of  the  use  of  these  formulas  may  be  found  in  the 
First  Course,  pp.  223-224. 

The  graphical  representation  of  all  the  solutions  of  one  equation  in 
two  unknowns  and  the  graphical  solution  of  a  system  of  two  equations 
in  two  unknowns  may  be  reviewed  at  this  point  if  desired.  (See  First 
Course,  pp.  170-173,  207-213,  232-234.) 


EQUATIONS 

„   ,                                 WRITTEN 
Solve : 

EXERCISES 

1. 

x  +  y  =  5, 

a;  -  ?/  =  3. 

11. 

4  x  —  3  y  =  3, 
3  a;  -  4  y  =  -  3. 

^2. 

a;  +■  y  =  5, 
a;  -  y  =  J . 

12. 

4:X  +  2y  =  l, 
3x-2y=*. 

3. 

2x-  +  y  =  3, 
a;  +  y  =  2. 

13. 

12  a;  - 11  y  =  87 

4  a;  +  2  ?/  =  46 

^4. 

a;-y  =  l. 

14. 

7  a;  -  2  ?/  =  3, 
7a;-4?/  =  — 1. 

5. 

2x  +  2y  =  8, 

2x-y  =  2. 

15. 

9  cc  —  3y  =  —  6, 

8  a;  -  2  y  =  -  6. 

6. 

3x  —  y  =  —  5, 
2x-y  =  -3. 

16. 

oa;  +  y  =  l, 

6a;  +  y  =  2. 

—■7. 

4x-3y  =  7. 
3x-4y  =  7. 

17. 

ax  -\-by  =  c, 
px  +  qy  =  tf. 

8. 

5x  +  y  =  9, 
3  x  -f-  y  —  5. 

18. 

a;  —  m?/  =  a, 
a;  +  py  =  b. 

^,9. 

4  a;  +  5  y  =  22, 
3  x  +  2  y  =  13. 

19. 

ax  —  by  =  e, 
ex  —  dy  =  f. 

10. 

a;  -  5  y  =  -  22, 

5#  —  y  =  10. 

20. 

ax  —  y  —  b, 
cx-\-y  =  d. 

33 


EQUATIONS  WITH  THREE  OR  MORE   UNKNOWNS 

77.  The  definitions  and  methods  for  the  solution  of  two 
equations  with  two  unknowns  may  be  applied  equally  well  to 
a  greater  number  of  equations  and  unknowns. 

To  solve  three  linear  equations  with  three  unknowns,  elimi- 
nate one  unknown  from  any  pair  of  the  equations  and  the 
same  unknown  from  any  other  pair;  two  equations  are  thus 
formed  which  involve  only  two  unknowns  and  which  may  be 
solved  by  methods  previously  given. 

Four  or  more  equations  with  four  or  more  unknowns  may 
be  solved  similarly. 


34  ELEMENTARY   ALGEBRA 

WRITTEN  EXERCISES 
Solve  and  test : 

1.  2x  +  3y  =  19,  7.   5x  =  3y, 
3x  —  4:y  =  3.  2x  +  Sy  =  ±. 

2.  5x-2y  =  46,  8-   ax-by  =  c,' 

x  + y  =  12.  cx  +  ay  =  b. 

3.  8x  +  7y  =  ll,  9-   4a2.«  +  5m/  =  35 
2x-3y  =  17.  6ax  +  7y  =  2. 

10.    ax  +  by  =  c, 

4.  4  a,*  —  by  =  l,  2     i  1,2         2 

^  era:  +  6-?/  =  c\ 

3  x  -  5  y  =  2. 


5.    7  a;  4- 9?/ 4- 1  =  0, 


11.   a?  4- &?/=!, 


3^  +  4^4  =  0.  a     V~   ' 

6.   17  a;  4-13  j/- 5  =  0,  12.  fa  +  |?/  =  i, 

3aj  +  6y  +  3  =  0.  f*  +  T762/  =  i- 

13.  2(2x  +  3y)-5^  +  3)-^  =  9,      a?  +  y  =  l. 

8  4 

14.  4?/  +  3(j/-a;-2)4-20a-  =  0, 
3(,_5)-2(.C-|)+2J  =  0. 

15.  4 a- 2 i/4-2  =  3,                   18.  3x -2  2  +  5  =  0, 
a  4-  3  ?/  4-  2  z  =  13,  2x4-3?/- 21=0, 
-8x  +  12y  +  z  =  21.  4 1/  +  7  2 - 69  =  0. 


16. 

5  a;  4- 4  y  +  2  2  =  17,                19-    »  +  y==-i 

3^-22/4-52  =  2,                                    a 
2a;-2/4-32  =  2.                          2/  +  z  =  i> 

0 

17. 

x  —  y  —  z  =  a,                                                1 
3  t/  —  a;  —  z  =  2  a,                            »  +  2  =  -  • 
72  —  y  —  x  =  4  a. 

20. 

•  +  y  +  z  =2c, 

a  +  b      b  —  c     a  +  c 

x         y         z    -*a 

a  —  b      6  —  c      a  —  c 

*        ,        V                2            o           o 

2 =  2  a  —  2  c. 

a  —  6     c  —  ?>     «  +  o 


EQUATIONS  35 

21.    x  —  y  —  z  —  2v:=  —  12, 
3x~y-2z  +  8w  =  4Q, 

4  x  —  4  ?/  +  7  z  —  5  w  =  52, 
3  a:  —  y  +  2  z  +  io  =  44. 

22.  A  merchant  bought  a  certain  number  of  platters  for 
$  366.  Three  were  broken  during  shipment.  He  sold  \  of  the 
remainder  at  a  profit  of  25%,  for  $75.  Find  the  number  of 
platters  bought  and  the  price  per  platter. 

23.  A  certain  hall  contains  both  gas  jets  and  electric  lights. 
When  60  gas  jets  and  80  electric  lights  are  used,  the  cost  for 
an  evening  is  $4.  If  90  gas  jets  and  60  electric  lights  are 
used,  the  cost  is  $  4.05.  Find  the  cost  per  gas  jet  and  electric 
light. 

24.  A  tailor  paid  $12  for  4  yd.  of  cloth  and  8  yd.  of  lining. 
At  another  time  he  paid  $21  for  6  yd.  of  the  cloth  and  16  yd. 
of  the  lining.     Find  the  price  of  each  per  yard. 

25.  A  certain  train  runs  25  mi.  per  hour  on  the  level,  15  mi. 
per  hour  on  up  grade,  and  30  mi.  per  hour  on  down  grade.  It 
goes  from  A  to  B,  200  mi.,  in  8  hr.  48  min.,  and  from  B  to  A 
in  9  hr.  12  min.  How  many  miles  are  level,  up  grade,  and 
down  grade  respectively  between  A  and  B  ? 

26.  Two  wheelmen  are  328  ft.  apart  and  ride  toward  each 
other.  If  A  starts  3  seconds  before  B,  they  meet  in  14  seconds 
after  A  starts ;  or  if  B  starts  2  seconds  before  A,  they  meet  in 
14  seconds  after  B  starts.     Find  the  rate  of  each. 

27.  A  man  had  a  portion  of  his  capital  invested  in  stocks 
paying  6  c/0  dividends,  the  remainder  in  mortgages  paying  5  %. 
His  annual  income  was  $  700.  The  next  year  the  dividend  on 
the  stock  was  reduced  to  5  %,  but  by  reinvestment  he  replaced 
his  old  mortgages  by  new  ones  paying  51%.  His  income  for 
this  year  was  $  690.  How  much  had  he  invested  in  stocks ; 
also  in  mortgages  ? 

28.  The  sum  of  the  digits  in  a  certain  number  of  two  figures 
is  13,  and  if  the  result  of  multiplying  the  tens'  digit  by  \\ 
is  added  to  the  number  itself,  there  results  a  number  with  the 
same  digits  in  reverse  order.     Find  the  number. 


36  ELEMENTARY  ALGEBRA 

QUADRATIC  EQUATIONS 

78.  Quadratic  Equations.     Equations  of  the  second  degree 
are  called  quadratic  equations. 

A  general  form  for  quadratic  equations  in  one  unknown  is 

ax2  +  bx  +  c  =  0, 

in  which  a,  b,  c  represent  any  known  numbers,  except  that  a 
may  not  be  zero. 

79.  Solution    of    Quadratic    Equations.     (1)  The    incomplete 
quadratic  equation  x2  =  a  is  solved  by  extracting  the  square 

root  of  both  members.     The  roots  are  :    x  =  ±  Va. 

(2)  The  incomplete  quadratic  equation  ax2  +  bx  =  0  is  solved 

by  factoring.     The  roots  are  x  =  0  and  x  = 

a 

(3)  Complete  quadratic  equations  are  solved  by  completing 
the  square. 

The  process  consists  of  two  main  parts : 

(a)  Making  the  left  member  a  square  while  the  right  member 
does  not  contain  the  unknown. 

This  is  called  completing  the  square. 

It  is  based  upon  the  relation  (x  +  a  )2  =  x2  +  2  ax  +  a2,  in  which  it  ap- 
pears that  the  last  term,  a2,  is  the  square  of  one  half  of  the  coefficient  of  c. 

(&)  Extracting  the  square  roots  of  both  members  and  soloing  the 
resulting  linear  equations. 

Square  roots  which  cannot  be  found  exactly  should  be  indi- 
cated. 


EXAMPLE 

Solve : 

^  _  8  x  +  9  =  0. 

(0 

Transposing, 

a;2_8a;=-0. 

(2) 

Completing  the  square, 

x2-8x+  16  =-9 +  10. 

(3) 

Rearranging, 

(x-4)2  =  l. 

(4) 

Extracting  the  square  root, 

x  —  4  =  ±  V7. 

(5) 

Solving  (/;)  for  x, 

x  —  4  ±  V7. 

(6') 

EQUATIONS  37 

(4)  If  any  quadratic  equation  has  zero  for  the  right  member, 
and  if  the  polynomial  constituting  the  left  member  can  be  fac- 
tored, the  quadratic  is  equivalent  to  two  linear  equations  whose 
roots  can  readily  be  found.     (See  First  Course,  pp.  186-187.) 

ORAL   EXERCISES 
Solve : 

1.  «2  =  16.  13.    x2  +  2  a; +  1  =  0. 

2.  r  =  64.  14.  y2-2y  +  l=0. 

3.  z2  =  8.  15.  z2-  3  2  +  2  =  0. 

4.  x2  =  a2.  16.  3x2  =  6x. 

5.  if  =  a2b2.  17.  (x-  l)(x  -  2)  =  0. 

6.  2  a?  =  8.  18.  x2-  5  £  +  6  =  0. 

7.  3  f  =  27.  19.  (x  -  7)(as  +  1)  =  0. 

8.  D22  =  125.  20.  x2  +  a;  +  \  =  0. 

9.  3*2  =  75.  21.  s4-16  =  0. 

10.  x(x-l)  =  0.  22.    t2-t  +  \  =  0. 

11.  a2  +  a;  =  0.  23.    £  _p2+ j»  + 1  =  0. 

12.  ?/2-4  =  0.  24.    z(ar-2z-3)  =  0. 

WRITTEN   EXERCISES 
Solve: 

1.  3  or  =  18.  3.    x2-5a;  +  6  =  0.        5.    t2-2t-6  =  Q. 

2.  x2  —  5x  =  Q.  4.    ar9+4a;-3  =  0.        6.    Sjpr=zop. 

7.  ar9  +  ll.i'  +  24  =  0.  s  =      9      .2x     ±35 

•   4      »- 8"      5 

8.  15x2-13x  +  5  =  0. 

11.   -J_+^L_ ±-  =  o. 

9.  15/ +  134?/ +  288=0.  y-3    1-y    y-2 

12.    7(7  -  z)(z  -  6)  +  3(5  -  2) (2  -  2)  -  40  =  0. 


13. 


6  —  2  mj     5  +  mj      w  —  5 


^+; 


io  —  2       3  +  (o      2  —  to 


38  ELEMENTARY   ALGEBRA 

14.  Find  two  numbers  whose  sum  is  10,  and  the  sum  of 
whose  squares  is  68. 

Suggestion.     Let  x  represent  one  number  and  10  —  x  the  other. 

15.  A  room  is  3  yd.  longer  than  it  is  wide ;  at  $  1.75  per 
square  yard,  carpet  for  the  room  costs  $49.  Find  the 
dimensions  of  the  room. 

16.  A  man  bought  for  $300  a  certain  number  of  oriental 
rugs  at  the  same  price  each.  If  he  had  bought  rugs  each 
costing  $40  more,  he  would  have  obtained  2  fewer  rugs.  How 
many  rugs  did  he  buy  ? 

17.  A  dealer  bought  a  number  of  similar  tables  for  $153. 
He  sold  all  but  7  of  them  at  an  advance  of  $  1  each  on  their 
cost,  thus  receiving  $  100.     How  many  tables  did  he  buy  ? 

18.  A  man  invested  $6000  at  a  certain  rate  of  simple 
interest  during  4  years.  At  the  end  of  that  time  he  reinvested 
the  capital  and  the  interest  received  during  the  4  years,  at  the 
rate  of  interest  1  °J0  lower  than  at  first.  His  annual  income 
from  the  second  investment  was  $  372.  What  was  the  original 
rate  of  interest? 

19.  A  rectangle  of  area  84  sq.  in.  is  5  in.  longer  than  it  is 
wide.     Find  its  dimensions. 

20.  A  certain  number  of  men  hire  an  automobile  for  $156. 
Before  they  start,  two  others  join  them,  sharing  equally  in  the 
expense.  The  amount  to  be  paid  by  each  of  the  original  hirers 
is  thus  reduced  by  $13.     How  many  men  were  there  at  first'.' 

21.  A  man  rows  down  a  stream  a  distance  of  21  mi.  and 
then  rows  back.  The  stream  flows  at  3  mi.  per  hour  and  the 
man  made  the  round  trip  in  13i  hours.  What  was  his  rate  of 
rowing  in  still  water  ? 

22.  The  product  of  a  number  and  the  same  number  increased 
by  40  is  11,700 ;  what  is  the  number  ? 

23.  If  each  side  of  a  certain  square  is  increased  by  5  the  area 
becomes  64  ;  what  is  the  length  of  a  side  ? 

24.  Find  two  numbers  whose  sum  is  16  and  the  difference 
of  whose  squares  is  32. 


EQUATIONS  39 

REVIEW 

WRITTEN   EXERCISES 

Solve : 

1     a  +  3_z-l  4.   2 x  +  3(4 x -1)  =  5(2 x  +7). 


aj  —  1      a;  +  3 

2.    ar -14 a;  +  33  =  0.  2a; 


~  =  6. 


3.    a  +  -=c.  6.   14aj-5(2a;+4)  =  3(aj+l). 

x 

1..    (5x-  2)(6  a;  +  l)-(10a;  +  3)(3  a  +  10)  =  0. 
8.    (2  x -  3)(x  +  1)  -  (3  x  -  7)(x-  -  4)  =  36. 

-        1     +_^=-^_.        12.    *=7, 


a;  +  2      x-2     2x  —  S  # 

2a?-10y  =  3y  +  2. 


10.    2av+5y  =  6, 

4x'  +  ll?/  =  3.  13. 


7y-6_l 
3a  +  4     2' 


11.    5  a;  — 3y  =  l,  4y  —  5_1 

13a;-8y  =  9.  2a;  +  l_3' 

,.     3x-  +  2i/  +  5     a?-3y-13  ,  y-3a;  +  3      Q 
14'  2  +  5  +  '" 6  =S' 

2  x  —  4  ?/  +  6  _    _r> 
?/  —  5  a;  +  11  " 

15.    6z  =  43-5y,  16.    ?~2=-, 

3Z  =  37-4a?,  *  +  ?     5 

.    4y  =  55-5a;.  '        *=*  =  * 

?/  +  z      3 

17.  3a;  +  4y  +  2s  =  -4,  »-y_l 
2  a?  -5  y-  z  =  9,                            ^Tz~1' 
-4a;  +  2y  +  3«  =  -23. 

18.  («-2)(a;  +  3)  =  (a;-l)(z-l), 

(z  +  8)(y-2)-(y  +  2)(z.+  2)  =  0, 
y(3-2aj)  +  (2y-3)(l+aj)  =  0. 


19. 


3  x2  + 14  ,  1  +  3  a;      3a;_n 


2  a; +  10      7(4  -aj)       5  + a;         7 


40  ELEMENTARY   ALGEBRA 

20.  In  a  certain  election  36,785  votes  were  cast  for  the 
three  candidates  A,  B,  C.  B  received  812  votes  more  than 
twice  as  many  as  A ;  and  C  had  a  majority  of  one  vote  over 
A  and  B  together.     How  many  votes  did  each  receive  ? 

21.  In  a  certain  election  there  were  two  candidates,  A  and 
B.  A  received  10  votes  more  than  half  of  all  the  votes  cast. 
B  received  4  votes  more  than  one  third  of  the  number  received 
by  A.     How  many  votes  did  each  receive  ? 

22.  A  group  of  friends  went  to  dine  at  a  certain  restaurant. 
The  head  waiter  found  that  if  he  were  to  place  five  persons 
at  each  table  available,  four  would  have  no  seats,  but  by  plac- 
ing six  at  each  table,  only  three  persons  remained  for  the  last 
table.  How  many  guests  were  there,  and  how  many  tables 
were  available  ? 

23.  A  flower  bed  of  uniform  width,  is  to  be  laid  out  around 
a  rectangular  house  20  ft.  wide  and  36  ft.  long.  What  must 
be  the  width  of  the  bed,  in  order  that  its  area  may  be  one 
third  of  that  of  the  ground  on  which  the  house  stands? 

24.  Wood's  metal,  which  melts  in  boiling  water,  is  made 
up  of  one  half  (by  weight)  of  bismuth,  a  certain  amount  of 
lead,  half  that  much  zinc,  and  half  as  much  cadmium  as  zinc. 
How  much  of  each  in  100  lb.  of  Wood's  metal  ? 

25.  If  in  the  preceding  exercise  three  fourths  as  much  cad- 
mium as  zinc  is  used,  a  metal  is  formed  that  melts  at  a  still 
lower  temperature.  How  many  pounds  of  each  constituent 
metal  are  there  in  100  lb.  of  this  metal  ? 

26.  A  cask  contains  10  gal.  of  alcohol.  A  certain  number 
of  quarts  are  drawn  out ;  the  cask  is  then  filled  up  with  water 
and  the  contents  thoroughly  mixed.  Later  twice  as  many 
quarts  are  drawn  out  as  the  previous  time  and  the  cask  filled 
up  with  water.  There  now  remains  only  4.8  gal.  of  alcohol 
in  the  mixture.  How  many  gallons  were  drawn  out  the  first 
time? 

27.  A  certain  fraction  not  in  its  lowest  terms  has  the  value 
\.  If  the  numerator  is  diminished  by  2,  and  the  denominator 
increased  by  2,  the  value  of  the  resulting  fraction  is  the  same 


EQUATIONS  41 

as  that  which  results  when  the  numerator  of  the  given  frac- 
tion is  doubled  and  the  denominator  multiplied  by  5.  Find 
the  fraction. 

28.  A's  investments  amount  to  $4000  less  than  B's,  and 
C's  amount  to  $6000  more  than  B's.  A's  average  rate  of 
annual  income  from  his  investments  is  one  half  of  1  %  more 
than  B's,  and  C's  is  one  half  of  1  %  less  than  B's,  and  A's 
annual  income  is  $80  less  than  B's,  and  C's  annual  income 
is  $120  more  than  B's.  Find  the  amount  of  each  man's 
investments,  and  each  man's  average  rate,  of  income. 


CHAPTER   III 

RADICALS 

DEFINITIONS   AND    PROPERTIES 

80.  Rational  Numbers.  Integers  and  other  numbers  expres- 
sible as,  the  quotient  of  two  integers  are  called  rational  numbers. 

Thus,  5  and  .2,  which  is  expressible  as  j%,  are  rational  numbers. 

81.  Irrational  Numbers.  Any  number  not  rational  is  called 
an  irrational  number. 

Thus,  \/2,  V3,  VlO,  — ,  1+V3,  V2— \/3, 7r,  are  irrational  numbers. 
V5 

82.  An  indicated  root  of  any  number  is  called  a  radical. 

Thus,  V5,  y/8,  \-r~i  ^a  + x2'  are  radicals. 

In  the  present  chapter  all  roots  that  cannot  be  exactly  extracted  by 
inspection  ars  indicated.  Methods  for  finding  approximate  numerical 
values  of  certain  roots  are  given  later. 

83.  Surd.  An  irrational  number  that  is  an  indicated  root  of 
a  rational  number  is  sometimes  called  a  surd. 

Thus,  V2,  Vo,  \/T,  are  surds. 

84.  An  expression  involving  one  or  more  radicals  is  called 
a  radical  expression. 

Thus,  5  +  2V3,  -^-  —  1,      +  V  _a,  are  radical  expressions. 
y/x  .  2  -  V3  b 

85.  Some  Properties  of  Radicals.  A  few  important  proper- 
ties of  radicals  are  given  here.  The  fuller  treatment  is  con- 
tained in  the  chapter  on  Exponents. 

42 


RADICALS  43 

86.    I.    Vo  •  V&  =  Va6. 

For  example,  y/2  •  V3  =  V6. 

That  this  is  true  may  be  seen  by  squaring  both  members. 

Thus,  (  V2  •  V3)  (  V2  ■  V3)  =  V6  ■  a/6, 

or,  V2  •  a/2  •  a/3  •  a/3  =  a/6  •  a/6, 

or,  2-3  =  6,  which  is  known  to  be  true. 

In  the  same  way,  it  may  be  seen  that  for  every  a  and  ft,  Va  •  Vb  =  Vab. 

In  words : 

The  product  of  tivo   square  roots    is   the  square  root  of  the 
product  of  the  numbers. 

WRITTEN    EXERCISES 
Show  by  squaring  that : 


1.  V3  •  V5  =  a/15.  5.  V2a  •  V3&  =  V67T&. 

2.  a/4  •  a/7  =  V28.  6.  Va?  •  V5y  =  a/5 afy. 

3.  a/3  •  \/7  =  V2T.  7.  V2  ■  V3  •  a/5  =  a/30. 

4.  VB  •  a/IT=  a/55.  8.  Vet  •  V&  •  Vc  =  V«6c. 

87.    II.    V^6=  Va2  Vb  =  ay/b. 

In  words : 

Square  factors  may  be  taken  from  under  the  radical  sign. 

Thus,  a/18  =  V9  •  a/2  =  VP  .  V2  =  3  V2. 

WRITTEN    EXERCISES 

Take  all  square  factors  from  under  the  radical  sign : 


1.  V20.  ■      5.  a/45.  9.  a/12.  13.  V8~a2. 

2.  a/27.  6.  V75.  10.  a/40.  14.  Vrf. 

3.  a/50.  7.  a/24.  11.  a/500.  15.  V48^y. 

4.  a/48.  8.  a/32.  12.  a/128.  16.  V46aV. 


44  ELEMENTARY   ALGEBRA 

88.    III.    a Vb  =  Va2  ■  V&  =  Va26. 

In  words : 

Any  factor  outside  the  radical  sign  may  be  placed  under  the 
radical  sign  provided  the  factor  is  squared. 

For  example : 

3  V2  =  V9  •  V2  =  Vl8. 

WRITTEN    EXERCISES 
Place  under  one  radical  sign  : 

1.  6V2.         5.   3-V7-2.  9.   4V2-3.      13.   tVg. 

2.  5V3.         6.   5.V3-V2.      10.    &V2.  14.   rVirr. 

3.  2-V3.       7.   2.V3-VH.    11.   2xVS^c.     15.   fVl8ay. 


4.   5-V7.       8.    5-V3-V7.      12.    a6V6c.       16.   aV&-a. 

PROCESSES 

89.  Preparatory. 

Read  and  supply  the  blanks : 

1.  3a  +  2a=(     -)a.     Similarly,    3V2  +  2V2  =  (     )V'2. 

2.  5a-3a=(     )a.     Similarly,    5V2-3V2  =  (     )V2. 

3.  7V3  +  3V3  =  (    )V3.         4.   8V5-6V5=(    )V5. 
5.    V75-  V12  =  5V3-2V3  =  (    )V3. 

90.  Addition  and  Subtraction  of  Radical  Expressions.  Radi- 
cals can  be  united  by  addition  or  subtraction  only  when  the 
same  root  is  indicated  and  the  expressions  under  the  radical 
sign  are  the  same  in  each. 

When  the  expression  cannot  be  put  into  this  form  the  sum 
or  the  difference  can  only  be  indicated. 


RADICALS  45 

91.    To  add  or  subtract  radical  expressions  having  the  same 
radical  part,  add  or  subtract  the  coefficients  of  their  radical  parts. 

For  example : 

2  V3  +  3  V3  =  5  V3. 

2  Vl2  +  V300  =  4  VS  +  10  V3  =  14  V3. 

Add  y/2,  -  V8,  \Vl6,  \V^54  : 

-V8=-2V2;    ¥l6  =  2</2;   v^54  =  -  3  y/2. 

.'.  V2  -  V8  +  \Vl6  +  V^54  =  v/2-2v/2  +  2\/2-3v/2 

=  _  V2-^. 

If  the  numerical  value  of  the  sum  or  the  difference  is  needed,  it  can  be 
found  approximately  by  methods  given  later. 

WRITTEN   EXERCISES 
Find  the  sum  : 

1.  V2,  V8,  VI8.  8.  V6,  V24,  V63. 

2.  V75,  -  V12,  -  V3.  9.  V108,  -  Vl2,  V48. 

3.  V8,  V5,  -  V18.  10.  Vt5,  V48,  -  V27. 

4.  V128,  -2V50,  V72.  11.  V80,  V20,  -  \/45. 

5.  -^40,  _  J/320,  ^135.  12.  VS,  -  V99,  Vl21. 

6.  8  V48,  - 1  Vl2,  4  V27.  13.  5  V24,  -  V54,  3  V96. 


7.    V72,  -3V9,  6V243.  14.    V27x4,  -V64x4,  VlGoA 

92.  Multiplication  of  Radical  Expressions  containing  Square 
Roots.  In  multiplying  expressions  containing  indicated  square 
roots,  make  use  of  the  relation  Va  •  V&  =  V«&. 

EXAMPLES 

1.     3-4V5  2.     2-    V3 

6  +  2V3  5  +  2V3 


18-24V5  10-5V3 
6V3-8VT5.  4V3-6 

18-24v/5  +  6v/3-8Vl5.  10-    V3  -  6  =  4  -  a/3. 


4G  ELEMENTARY   ALGEBRA 

,  ,  .    ,  WRITTEN    EXERCISES 

Multiply : 

1.  2  +  V5by2-V5.  7.    4  +  V5byVlO. 

2.  1+V3by2+V5.  8.    3-Vl5by2+VK 

3.  2+V3by2  +  V3.  9.    1+V2byl-VB. 

4.  V2  +  V3  by  1  -  a/3.  10.   2  V3  -  3  v'5  by  V3  -  VS. 

5.  V3-V5by  V3  +  V5.  11.    V14  +  V7  by  V8  -  V21. 

6.  Vf>  -  V6  by  V5  -  V6.  12.    V5  -  a/48  by  a/5  +  Vl2. 

93.    Division   of   Square   Roots.     The  quotient  of  the  square 
roots  of  two  numbers  is  the  square  root  of  the  quotient  of  the 

Va  _      ja 
numbers.     In  symbols,  —=  —  \  ,  " 


V5       .  |5 


Thus,  ^  =  a  ,  - ,  because,  multiplying  each  member  by  itself, 
V6       ^ 


V6     V6 

V0-V6       >6      ^ 


or 


or  -  =  -  ,  which  is  evidently  true. 

0      6 


ORAL    EXERCISES 

Read  each  of  the  following  as  a  fraction  under  one  radical 
sign: 

1.    VS.  3.  ^*.  5.   -!_  7.   -^t 

Vo  V7  V5  VIS 

2  vi.        4.  j_=yr.       6.  vs.  8  yjo. 

V7  V2      V2  V8  V20 


RADICALS  47 

94.  Rationalizing  the  Denominator.  Multiplying  both  numer- 
ator and  denominator  of  a  fraction  by  an  expression  that  will 
make  the  denominator  rational  is  called  rationalizing  the  de- 
nominator. 

Thus,  multiplying  both  numerator  and  denominator  of by  V'2,  we 

obtain  "^ 

V3  _  V2  •  V3  _  V6 

V2       V2  •  V2        2 


WRITTEN    EXERCISES 
Rationalize  the  denominator  of: 


1.  1 . 

V2 

4.   2- 
V3 

7  V5 
V7 

10.  10 

V5 

2.  JL. 

V3 

5.  3- 

V7 

8.  A. 
V3 

ii.  A 
V3 

3.  J-. 

o. • 

V3 

9.  A 

V3 

12.  A 

V2 

95.  Rationalizing  Factors.  When  the  denominator  is  of  the 
form  Va  +  V&  or  a  +  V&,  the  rationalizing  factor  is  the  same 
binomial  with  the  connecting  sign  changed,  often  called  the 
conjugate  binomial. 

It  is  not  necessary  in  elementary  algebra  to  take  up  the  rationalizing 
of  more  complicated  denominators. 


EXAMPLES 

3 

1.    Rationalize  the  denominator  in 


2-V5 
The  conjugate  of  2  —  Vh  is  2  +  Vb. 

Then,        3      •=■ 3(2+V5) J  +  3V5  =  _  (fl  +  3^ 

2  -  Vo      (2  -  V5)  (2  +  V5)         ^  -  5 


48  ELEMENTARY   ALGEBRA 

2  +  V3 


2.   Rationalize  the  denominator  in 


V3  +  V5 
The  conjugate  of  V3  +  Vb  is  V3  —  VE. 

Then  2+V3  (2  +  \/3)(V3  -VI)    _  2>/3  +  3  -  2V5  -  VT5 

V3+V5      (•v/3  +  v/5)(V3-V'5)  3~5 

3  +  2V3-2V/5-V15 

~~  2 

WRITTEN     EXERCISES 

Rationalize  the  denominators : 

!     2  +  VF.  5  5  9    3+V5 

3+V3*  V3+V7  3-V5 

3V3  +  2V2.  6     2-Vg  1Q     8-5V2 

V3-V2  3-V5  3-2V2 


1 


3  n     2  +  4V7 


1_V2  V5  +  V2  2V7-1 

3  g     V5  +  2V2,  12      2VT5-6 


2-V5  "      4-2V2.  V5+2V2 

RADICAL  EQUATIONS 

96.  To  solve  eqxiations  in  which  only  a  single  square  root 
occurs,  transpose  so  that  the  square  root  constitutes  one  member. 
Square  both  members  and  solve  the  resulting  equation. 

EXAMPLE 


Solve:  2x-3=Vx2  +  6x-6.        (2) 

Squaring  both  members,  4  X2  —  12  X  +  9  =  X2  +  6  X  —  6.  (£) 

Collecting  terms,  3  X2  —  18  X  +  15  =  0.  (5) 

Dividing  both  members  by  3,        X2  —  6  X  +  5  =  0.  (4) 

Solving  (4),  x  =  5orl.  (5) 

Test.     On  trial,  it  appears  that  5  satisfies  the  given  equation,  taking  the 
radical  as  positive,  while  1  satisfies  the  equation  2  x—  3  =  —  Vx2  +  6  x  —  0. 


RADICALS  49 

1.  It  must  be  remembered  that  the  equation  resulting  from  squaring 
will  usually  not  be  equivalent  to  the  given  equation  (First  Course,  Sec. 
232,  p.  186).  It  may  have  additional  roots,  and  trial  must  determine 
which  of  the  roots  found  satisfy  the  given  equation. 

2.  In  order  that  the  given  problem  may  be  definite,  the  radical  must 
be  taken  with  a  given  sign.  If  every  possible  square  root  is  meant,  two 
different  equations  are  really  given  for  solution.     Thus,  unless  restricted, 

2  x  =  V4  —  6  x  is  a  compact  way  of  uniting  the  two  different  equations, 

2  x  =  +  V4  —  6  x,  and  2  x  =  —  V4  —  6  x.     If  solved  as  indicated  above, 
it  appears  that  the  first  is  satisfied  when  x  =  \,  the  second  when  x  =  —  2. 

3.  In  the  exercises  of  the  following  set  the  radical  sign  is  to  be  un- 
derstood to  mean  the  positive  square  root. 

WRITTEN    EXERCISES 

-      Solve: 

1.  x  =  Vl0x  +  7.  7.  30  =  x  -  29  Vx. 

2.  x  =  Vb  +  x  -  bx.  8.  x  =  2  +  V'S-llx. 

3.  3  x  -  7  Vx  =  -  2.  9.  x-  VaT+2  =  3. 

4.  Vx  +  5  -  x  =  - 1.  10.  8 x  +  1  =  VxT3«. 


Vr-1 


a; 


3  16 


11.  V100  -  x2  =  10 


x. 


6.  x  -(-  5  V37  -  x  =  43.  12.  x  +  V2  s  -  x2  =  6. 

SUMMARY 
I.    Definitions. 

1.  Rational    numbers  are  integers  and   other   numbers  ex- 
pressible as  the  quotient  of  two  integers.  Sec.  80. 

2.  An  irrational  number  is  any  number  not  rational. 

Sec.  81. 

3.  A  radical  is  an  indicated  root  of  a  number.  Sec.  82. 

4.  A  surd  is  an  irrational  number  that  is  an  indicated  root 
of  a  rational  number.  Sec.  83. 

5.  A   radical  expression   is   an  expression  involving  one  or 
more  radicals.  .  Sec.  84. 


50  ELEMENTARY   ALGEBRA 

II.    Properties  and  Operations. 

1.  The  product  of  two  square  roots  is  the  square  root  of  the 
product  of  the  numbers.  Sec.  86. 

2.  Square  factors  may  be  taken  from  under  the  radical  sign. 

Sec.  87. 

3.  Any  factor  outside  the  radical  sign  may  be  placed  under* 
the  radical  sign  provided  the  factor  is  squared.  Sec.  88. 

4.  Radical  expressions  whose  radical  parts  are  the  same  may 
be  added  or  subtracted  by  adding  or  subtracting  the  coefficients 
of  their  radical  parts.  Sec.  91. 

5.  The  quotient  of  two  square  roots  is  the  square  root  of  the 
quotient  of  the  numbers.  Sec.  93. 

6.  Multiplying  both  numerator  and  denominator  of  a  frac- 
tion by  a  factor  that  will  make  the  denominator  rational  is 
called  rationalizing  the  denominator.  Sec.  94. 

7.  Radical  equations  involving  only  a  single  square  root  are 
solved  by  transposing  so  that  the  square  root  constitutes  one 
member,  squaring  and  solving  the  resulting  equation.     Sec.  96. 


REVIEW 

Simplify : 

WRITTEN    EXERCISES 

1. 

V5 
V60 

4.     —  -  • 

V40 

2. 

v     8             v  8" 

5.    V50+V128. 

3. 

V6  •  V125. 

6.    VS-V5. 

7>    3V3  +  2V2 
V3-V2 

8.  V6h-V2. 

9.  (2  +  V3)2. 

10.    (5  +  V7)(5-V7).  11.    (2  V3  +  3V5)WI5. 

12.    (V6  + V15)(V8-V20). 
Solve : 


13.    Vx+5==x  —  7. 


16.  x  +  ^Jx  +  5  —  2x  —  1. 


14.  V2.r  +  7  =  |.      17.  .-2  +  V2~;=0. 

,e    a-10  .  x-1  ,  V2.X--1      .. 

15.    —• =U 


2  3  2  18.  x+Vd-x-=A. 


RADICALS  51 


SUPPLEMENTARY    WORK 

ADDITIONAL   EXERCISES 

Express  with  rational  denominators,  and  with  at  most  one 
radical  sign  in  the  dividend : 


1. 

•V12-5-  V3. 

2. 

V7- vn. 

3. 

2V24-5-2V6. 

4. 

2  -=-  3  V5. 

X, 

4 

V5-1 

6. 

l-:-(V2-10).' 

7. 

V2  -  (V2  -  V3). 

8. 

(2V6  +  5V12)-  V6. 

9. 

(5  V18  -  8  V50)  -5-  2  V2 

10. 


11. 


12. 


13. 


14. 


8-5V2 
3-2V2* 

3  +  V5 
3- V5 

V3    _ 

V5-V3 

2.+  4V7 
2V7-1 

4V7+3V2 
5V2  +  2V7* 


Square  Root  of  Binomials  of  the  Form  a  +-  V6 

Binomials  of  the  form  a  +  V&  can  often  be  put  into  the  form 
x  +  y  +  2  %/.»?/,  or  ( V.c  +  Vj/)2,  and  hence  the  square  root, 
V*c  +  Vy,  of  the  binomial  can  be  written  at  once. 

EXAMPLES 

1.    Find  the  square  root  of  4  +  2V3. 

4  +  2VS  =  3  +  1  +  2V3TT. 
Hence,  a;  +  ?/  =.3  +  1  and  a;?/  =  3  -'1,  from  which  x  =  3  and  y  =  1. 

.-.  V4  +2V3=  ±  (V3  +  VI)  =  ±  (V3  +  1). 

The  coefficient  of  the  radical  must  be  made  2  in  order  to  apply  the 
formula  x  +  y  +'2\'xy. 


52  ELEMENTARY   ALGEBRA 

2.    Find  the  square  root  of  3  —  V8. 

3  -V8  =  3  -  V4T2  =  3  -  2  V2.  • 
.  ■.  x  +  y  =  3,  and  xy  =  2.     ,\  a;  =  2,  y  =  1  by  inspection. 


.-.  V3-V8=  ±(V2-Vl)  =  ±(\/2  -  1). 

3.    Find  the  square  root  of  7  +  4  V3. 

7  +4V3  =  7  +  2vTT3;  x  +  y  =  7,  xy  =  12;   .-.  x  =  4,  ?/  =  3. 


.-.  V7  +  4V3  =  ±  (VT  +  V3)  =  ±  (2  +  V3). 

The  square  root  as  a  whole  may  be  taken  positively  or  negatively,  as  in 
the  case  of  rational  roots. 

The  solution  of  these  problems  depends  upon  finding  two  numbers 
whose  sum  and  product  are  given.  This  can  sometimes  be  done  by 
inspection,  but  the  general  problem  is  one  of  simultaneous  equations. 
See  Sec.  168,  p.  132. 

WRITTEN    EXERCISES 
Find  the  square  root  of : 

1.  11  +  6V2.  4.    41  -  24V2.  7.    17  +  12V2. 

2.  8-2V15.  5.   21  -V5.  8.   IV5  +  31, 

3.  49  -  12 V10.  6.   2|  -  |V3.  9.   56  -  24 VE. 


CHAPTER   IV 

EXPONENTS 

LAWS   OF  EXPONENTS 

97.  Preparatory: 

1.  What  is  the  meaning  of  a2?  Of  «4?  Of  a7?  Of  a11? 
Of  a"  ? 

2.  What  is  the  meaning  of  Vo2?  Of  tya??.  Of  K/a™? 
Of  a/^'? 

3.  «2 .  «3  =  ?     a3  •  a2  =  ?     a5  •  a3  =  ?     a8  •  a5  =  ? 

4.  a3-j-a2=?     a4-s-a2=?     as-»-a2  =  ?     bw---bi=? 

5.  (a2)2  =  ?     (a2)3=?     (c5)2=?     (V°)3=? 

98.  We  shall  soon  define  negative  and  fractional  exponents, 
but  until  this  is  done  literal  exponents  are  to  be  understood  to 
represent  positive  integers. 

99.  Law  of  Exponents  in  Multiplication. 
I.  am  -  ar  =  am+r. 
For                   am  =  a  ■  a  ■  a  •  •  •  to  m  factors, 

and  ar  =  a  ■  a  ■  a  ■■■  to  r  factors. 

.-.  am  •  ar  —  (a  •  a  ■  a  •  ••  to  m  factors)(a  .a  •  a  •••  to  r  factors) 
=  a  ■  a  ■  a  ■  a  •  •  •  to  m  +  r  factors 
=  am+r,  by  the  definition  of  exponent.  . 
Similarly,         am  •  ar  ■  aP  •■■  -  am+r+P  •■-. 

Multiply:  0RAL  EXERCISES 

1.  a2-a\       4.    ml-m5.       7.    (-l)3-(-l)5.      10.    23  •  23  •  22. 

2.  a3-a\       5.    x>x.  8.    62  •  62.  11.    7  •  72  •  7 ■". 

3.  a5 -a7.       6.    23-24.         9.    5  •  5  •  52.  12.   3-35-32. 
13.    (_l)2.(-l)3.(-l)l           14.    (_«)2.  (-«)<•(-«). 


54  ELEMENTARY  ALGEBRA 

100.    Law  of  Exponents  in  Division. 

II.  —  =  am-r,  if  m  >  r. 

ar 

For       am  =  a  •  a  •  a  •■•  in  factors, 

and  aT  =  a  ■  a  ■  a  ■■■  r  factors. 

am _a  •  a  •••  m  factors 
ar       a  ■  a  ■■•  r  factors 

=  a  ■  a  ■  a  •••  in  —  r  factors,  canceling  the  r  factors  from  both 
terms 

=  am~r,  by  definition  of  exponent. 
Divide:  0RAL   EXERCISES 


1. 

a4 

a2' 

5. 

a8 

'a5 

9. 

63 
62' 

13. 

x*ym 

X 

2. 

a5 
a3' 

6. 

(-a)5, 
(-a) 

10. 

54 
53' 

14. 

V 

3. 

a7 
a5' 

7. 

(-1)5. 
(-1)3 

11. 

4r 

15. 

.     t 

4. 

24 
22' 

8. 

(a6)2 
(aft) 

12. 

X 

16. 

TT'fi 

r 

101.    Laws  of  Exponents  for  Powers. 

III.  (amy  =  amr. 

For  (am)r  =  am  •  am  •  am  ■••  to  r  factors 

=  (a  •  a  •■■  to  m  factors)(a  •  a  •••  to  m  factors)  to 
r  such  parentheses 

=  a  •  a-  a  ■■■  to  mr  factors 

=  a""',  by  definition  of  exponent. 

ORAL    EXERCISES 

Apply  this  law  to : 

1.  (43)2.  4.  («3)2.  7.  O2)5.  10.  [(ft)4]5. 

2.  (32)5.  5.  (a2)3.  8.  (a3)8.  11.  [(-a)5]2. 

3.  (-/.  6.   (a5)2.  9.   (</'/.  12.   [(-b/j;. 


EXPONENTS  55 

IV.  (ab)»  =  a"b". 

For  (ab)n  =  (nb)  •  (ah)  •••  to  n  factors 

=  (a  •  a  •••  to  n  factors)(6  •  b  •••  to  n  factors) 
=  a»l>'\  by  definition  of  exponent. 

Similarly,  (abc  •••)"  =  a"bnci  •••. 

ORAL    EXERCISES 

Apply  this  law  to : 

1.  (8  •  3)2.  4.  (ab)5.  7.  {mn)p.  10.  (afy)3. 

2.  (4  •  5)2.  5.  (cdf.  8.   (.>•>,)*.  11.  (Vy2)5. 

3.  (2-5)3.  6.  (abcy.  9.   (a&f.  12.  (aa8)6. 


V. 


?Y  — — 

6/    "6" 


(f)"= 


For  =  -  •  -  •  •  •  to  n  factors 


b     b 

a  •  a  ■■■  to  n  factors 
b  •  b  •••  to  n  factors 


=  — ,  by  definition  of  exponent. 
bn 


ORAL    EXERCISES 

Apply  this  law  to : 

i-  ay-  .  ftY.  12. 


(f- 


2-  (i)3-  ' 

3-  (I)5-  9.    fCX  13.    ^Y 


*•    (t)4' 


6 


5-    ("f)2-  in     W.  14.    U 


6-    (-f) 


3\3 


y 


5d 


7 


11.    W  15. 

n 


5G  ELEMENTARY   ALGEBRA 

102.  Collected  Laws  of  Exponents. 

I.    am  •  ar  =  am+r.  Sec.    99. 

II.   «m  -v-  ar  =  am-r.  (m  >  r.)'  Sec.  100. 

III.  (am)r  =  amr.  Sec.  101. 

IV.  (ab)n  =  a"bn.  Sec.  101. 

V    f«V  =  --  Sec.  101. 

\jbj       bn 

FRACTIONAL  EXPONENTS 

103.  Hitherto  we  have  spoken  only  of  positive  integers  as 
exponents,  the  exponent  meaning  the  number  of  times  the 
base  is  used  as  a  factor.  This  meaning  does  not  apply  to 
fractional  and  negative  exponents,  because  it  does  not  mean 
anything  to  speak  of  using  a  as  a  factor  f  of  a  time,  or  —  6 
times.  But  it  is  possible  to  find  meanings  for  fractional  and 
negative  exponents  such  that  they  will  conform  to  the  laws  of 
integral  exponents. 

104.  Preparatory. 

Find  the  meaning  of  a2. 

i      i        i+i 
Assuming  that  Law  I  applies,  a?  •  a%  =  a?  *  =  a, 

or,  (a*)2=  a. 

That  is,  a?  is  one  of  the  two  equal  factors  of  a, 

or,  a2  =  va. 

Thus,  the  fractional  exponent  \  means  square  root. 

ill        i+i+i 
Similarly,  a*  •  a*  •  a3  =  a3   3    3  =  a, 

i 
or,  (a3)3  =  a. 

That  is,      a3  is  one  of  the  three  equal  factors  of  a, 

or,  a*=  Va. 

Thus,  the  fractional  exponent  £  means  cube  root. 


EXPONENTS  57 


WRITTEN    EXERCISES 
Find  similarly  the  meaning  of : 


1.     dfl,                          3.     X*. 

5.    ri*.                   7. 

i 
vnJ. 

2.    </■■.                     4.    &A- 

6.    c*.                     8. 

a?\ 

1 
105.  The  meaning  of  a» 

is 

found  as  follows : 

Assuming  that  Law  I  applies 

» 

i      i       i 

an  •  a11  •  a"  •••  to  n 

factors 

-  -1 1 1-  -  •  •  to  n  terms. 

—  qU      n      n 

1 
n  •  — 

=  a     » 

»1 


=  a1  —  a. 

l  i 

That  is,  an  is  one  of  n  equal  factors  of  a,    or  a"  =  Va. 


106.  Preparatory. 

2 

Find  the  meaning  of  a?. 

2  2  2  fi  2 

Assuming  that  Law  I  applies,  a5  •  cfi  •  a*  =  a?  =  a2,     or  (a?)3  =  a2. 
That  is,  a?  is  one  of  the  three  equal  factors  of  a2,  or  aT=  Va2. 

p 

107.  The  meaning  of  aq  is  found  as  follows : 


Assuming  that  Law  I  applies, 


'-'       2       P  M  P  +  E  +  P...  to « terms 

a« .  a? .  a«  •••  to  g  factors  =  a «    «    « 

p 

9   ■  - 

—  a     «  =  a?. 

That  is,  a*  is  one  of  the  q  equal  factors  of  up, 

p        

or,  ai=VaP. 

P  -.  V 

Similarly,  •  ai  =a"  '  P  =  (Va)P. 

In  words : 

a  with  the  exponent  --  denotes  the  qth  root  of  the  pi/i  power  of 
a,  or  the  \>th  power  of  the  ([th  root  of&. 


58  ELEMENTARY   ALGEBRA 

This  definition  applies  when  p  and  q  are  positive  integers.     The  mean- 
ing of  negative  fractional  exponents  is  found  in  Sec.  116,  p.  64. 


ORAL    EXERCISES 


1.  State  the  meaning  of ; 


1.  53.                3.   6*. 

5.   4i 

7.   8*. 

2.    a*.                4.    ai 

6.    6*. 

8.    c\ 

Find  the  value  of : 

9.    8s.               io.    16*. 

11.   25* 

12.   32*. 

WRITTEN    EXERCISES 
Express  with  fractional  exponents : 

1.    </a*. 


2.    Va2. 


3  / * 

3.     vraw.  12 


4.  VaV&. 

5.  Va&. 


s/32  «568 

13.  ^33. 


6.    -Vatfxy.  e/64*5 

14. 


3/o;4w2 


>■  # 


.'/ 


15.    Va  +  &. 

\-— •  16.    Va3+63. 


9.    Vl6  ^.  17.    VP  •  Va. 


19. 

Va. 

20. 

V(«-6)2. 

21. 

Va-Vo. 

22. 

Va™ 

23. 

VcT». 

24. 

W\ 

25. 

Vam". 

26. 

2"Va2™. 

27. 

"Va»6'w. 

28. 

Va  •  Va. 

29. 

Vaoc. 

30. 

Va'  •  Va7 

10.    VmWVp.         18.    V-a-V-6. 

108.  The  definition  of  positive  fractional  exponents  has  been 
found  as  a  consequence  of  the  assumption  that  Law  I  applies 
to  them.  It  can  be  shown  that  the  other  laws  of  Sec.  102, 
p.  56,  also  apply  to  this  class  of  exponents,  as  thus  defined, 
and  we  shall  so  apply  them,  although  the  proof  is  omitted  here. 


EXPONENTS  59 

109.,   According  to  Law  I  (Sec.  99),  when  the  bases  are  the 
same,  the  exponent  of  the  product  is  found  by  adding  the  exponents. 

1  3  1.8  5 

For  example  :  a2  ■  a4"  =  a-   *  =  a4". 

A  general  formula  for  this  statement  is, 

m  p  mq     np  mq+np 

a"  •  a q  =  a"q   nq  —  a  m  • 
The  number  a,  or  the  base,  must  be  the  same  in  all  factors.     When  it 

1  3 

is  not,  as  in  a2  •  6¥,  the  product  cannot  be  found  by  adding  the  exponents. 


WRITTEN   EXERCISES 

Find  the  products : 

1.    a$  •  ai                     6. 

a'3  •  a* 

11. 

i       i 

xa  •  xb. 

2.    42-4l                      7. 

3               3 

m*  •  m3. 

12. 

m          p 

xn  •  xn. 

3.   7*.  7*.                    8- 

a^  •  a?. 

4.    a2"  •  ai                     9. 

i      i 

13. 

2            1 
pi    .  pZ  .  p. 

5.    &*  .  &i.                      io. 

1           1 

a"  •  a"1. 

14. 

4             3             1 

a5  •  a4  •  a2". 

110.  According  to  Law  II  (Sec.  102),  when  the  bases  are  the 
same,  the  exponent  of  the  quotient  is  found  by  taking  the  difference 
between  the  exponents. 


1  1  2_1  1 

For  example  :  a3  -s-  a2  =  a3  .2  =  a*. 


a  h-  a*  =  a    *  =  a*. 
A  general  formula  for  this  statement  is, 

m  p  m     p  mq    np 

an^-aq  =  a"   1=a  ng    . 


-  is  here  supposed  to  be  greater  than  -,   but  this  restriction  will 
n  q 

be  removed  later. 


GO  ELEMENTARY   ALGEBRA 

WRITTEN   EXERCISES 
Find  the  quotients : 


1. 

3                 1 

7. 

3                2 

12. 

2. 

a  -h  aT2. 
i 

8. 

2                1 

x5  -:-  a;7. 

13. 

i         i 

ma  -7-  mb. 

3. 

a  -i-  an. 
p 

9. 

a&-j-(a£)7. 

14. 

„2                1 

6* -=-6*. 

4. 

5. 

a  -i-  a7. 

4               3 

a5  -s-  a*. 

10. 

W    "  W 

15. 

3                   1 

m8  -7-  m' 

6. 

1                2 

XB  -i-Xs. 

11. 

•>        i 
53-=-52. 

16. 

4                3 

J9?H-p4. 

111.  According  to  Law  III  (Sec.  102),  when  an  exponent  is 
applied  to  a  base  having  an  exponent,  the  product  of  the  exponents 
is  the  exponent  of  the  result. 


i  2.1 

For  example  :  (a2)'2  =  a     2  —  ax  =  a. 

1  2.1  2  1_ 

(«2)4  =  a     ^  =  al  =a^. 

12  1.2  _2 

l3\~5  —  n.S      5  —  ,7  13. 


(a3)5  =  a5 


A  general  formula  for  this  statement  is, 

m   V-  m    p  mp 

(a»)q=an  '*  =  «"*. 


~.        ,.„  ORAL   EXERCISES 

Simphty : 

1.  (2*)*.  5.    (b^)K             9.    (at*)*.  13.  (a*)* 

2.  (3*)2.  6.    (a*)i            10.    0/9)*.  14.  («*)4. 

3.  (3*)3.  7.    (a^)i           11.    (6*)*.  15.  (10*)*. 

4.  (5*)2;  8.    (cfyk            12.    (3*)i  16.  [(a+&)£]& 
17.  [(a-byy.            18.    [K-6;!j'.  19.    [(.e"->/>> 


EXPONENTS  61 

112.    According  to  Law  IV  (Sec.  102),  an  exponent  affecting 

a  product  is  applied  to  each  factor,   and  according  to  Law  V 

(Sec.  102),  an  exponent  affecting  a  fraction  is  applied  to   both 

numerator  and  denominator. 

i       ii 
For  example  :  (ab)2  —  a2b^. 

(abh2)2'  =  ahh. 

(8  x6y2z) 3  -  8*x*ySz*  =  2  x2y^z^. 

m     p  mp       mp        p 

(ambnc)i  =  a  i  ■  b'">  •  c«. 


(ys)f     2/f 


A  general  formula  for  this  statement  is 

mr        '  pr 


5 

n 

mr        '    vr  ~ 

a" 


—     =  ans  h-  bqs ,  or 

p    J  >  in_ 

bq" 


WRITTEN   EXERCISES 


Simplify : 

1. 

(a?b*)k 

2. 

(arty. 

1 

3. 

(ambn)p. 

4. 

(x^y^y. 

5. 

(a36-f)2 

7.    ($Qx«y)i  ii.   (tt\. 

0    {32  a5bwy 

[~~^~J  '  12-    (16A8)*- 


/G4^12\*  13-    («6^915)3- 


y 


14.   (27a}Wy. 


i 


f9  aW 


6.    (a*  •  6*)8.  10"    V    «    7  '  IB-   (mW)M- 


62  ELEMENTARY  ALGEBRA 

113  When  the  bases  are  different  and  the  fractional  expo- 
nents are  different,  the  exponents  must  have  a  common  denomi- 
nator, before  any  simplification  by  multiplication  or  division 
is  possible. 

For  example  :  a~2b3  -  a?65  =  («^»2)*. 

A  general  formula  for  this  statement  is, 

m     p  mq  np  1 

a  "&«  =  aF9  •  b"9  =  (amq .  b"")"9 . 

This  is  called  simplifying  by  reducing  exponents  to  the  same 
order. 

WRITTEN    EXERCISES 

Simplify  by  reducing  the  exponents  to  the  same  order : 


1. 

1     ,i 
a2  •  b3. 

5. 

1         ,3 

a3  •  &*. 

9. 

lii 

2. 

2        .  3 

a3  •  &*. 

6. 

&*  •  6*. 

10. 

2            3               1 

m3  •  n5  -i-p*, 

3. 

i     ,  i 
■v*  •  b3. 

7. 

5*  •  &*. 

11. 

i       ii 

Wn  ■  Qm  •  It. 

4. 

i     ,  i 

8. 

i      ,  i 
a?  -=-  &*.■ 

12. 

p          m          1 

x9  •  yn  ■  zn. 

114.   It  is  usually  preferable  to  indicate  roots  by  fractional 

exponents  instead  of  by  radical  signs,  since  operations  are  thus 

more  easily  seen. 

COMPARISON 
By  Radicals  By  Exponents 

1.  Vffl y/a  =  VopVa?  =  y/a3a'z  —  VaK  a^a*  =  a*a*  =  a* +  6  =  0s. 

WRITTEN   EXERCISES 

Simplify  by  use  of  fractional  exponents  as  in  the  examples 
above : 

l.   22  •  5*.  3.    VSla^.  5.   3*  .2.7*. 


2.    Vy4^-.  4.    V5  •  VT5.  6.   3  •  5*  ■  2V3. 


EXPONENTS  63 


7.    V8-3V2.  21.    (16 x4jj)K  34    /49a4^ 

'  V04  66 


8.  2V3  -3  •  V10.  22     V32aa&10. 

9.  2a/3-3v/2.        oq        ,,,,.     4  2 

23.    Vob  7/r/r. 


35.    Va8V25&. 


36.   ^x2y^/2oyz. 


10.  52  •  3  •  5i  nA       ,— — — . 

24.  Vb4  mV,  6 

11.  V7  •  11*  '  .  37-   Va^- 

25.  (49a4^. 

12.  -8V2-12V3.  38.  V^Irf. 

26.  a/2o¥2.  _ 

13.  -VI2-2V3.  39.    V</8^. 

27.  V32  •  V2.  

14.  aVb-bVa.  40.   V^/sTa4. 

28.  (a2bVxJ2f. 


15.   (V2-V3)2V3.  41.   V^^9¥2. 

*,*      „    ^      29-    (m3^7)4' 


16.  3^(6^  -  2  •  5*).  42.  V^^/27  a3. 

3A-  30.    V6a&  •  V2a. 

17.  (a-Vxy.  43_  a(a26)^.6(a^)i 

,„       /-en  31.    [2a(4a2)*]2.                 t „ 

18.  Va668.  L                J       44.  3V9  a- 3*. 

19.  v8a°^fa.  45.  5  v a¥  •  2  Vacc. 


20.    (27  ay2)*.  33.  Vm2^/m!         46.  3 V8  •  2 ^6  •  3 </54. 

MEANING    OF    ZERO   AND    NEGATIVE    EXPONENTS 

115.   The  meaning  of  an  may  be  found  as  follows: 

Assuming  that  Law  I  holds  for  a0,        a5  •  a0  =  a5+0 

=  a5. 

a5 
Dividing  by  a5,  a0  =  —  =  1. 

«s 
That    is: 
Any  number  (not  zero)  with  the  exponent  zero  equals  1. 

Thus,  5"  =  1,    10°  =  1,    (Y)°  =  1, 


5. 

(-3)°. 

10. 

\  ■  a0. 

15. 

G)°- 

6. 

(*)•■ 

11. 

a5a°. 

16. 

.9  -s-  100n. 

7. 

(-^r)°- 

12. 

am  ■  b°. 

17. 

9i  •  5n. 

8. 

32  •  3n. 

13. 

a5  -=-  a0. 

18. 

OO      9J0 

9. 

32  •  5°. 

14. 

43--4°. 

19. 

(23  •  32)°. 

G4  ELEMENTARY   ALGEBRA 

„...,-        ORAL   EXERCISES 

State  the  value  oi : 

1.  (a b)°. 

2.  .9°. 

3.  100n. 

be 

116.    The  meaning  of  the  negative  exponent  may  be  found 
as  follows : 

Assuming  that  Law  I  holds  for  negative  exponents, 

5-3  .  5+3 --5-3+3- 50-1. 

That  is,  5~3  is  a  multiplier  such  that  its  product  with  5+3  is  1.    But  if 
the  product  of  two  numbers  is  1,  one  is  the  reciprocal  of  the  other. 

Therefore,  5-8  is  the  reciprocal  of  68  which  is  — . 

6 
Expressed  in  general  terms  : 

a~n  ■  an  -  «-"+"■ 

=  «°  =  1. 

In  words : 

a~n  means  —,  for  all  values  of  n,  positive  or  negative,  inte- 
an 
gral  or  fractional. 

ORAL  EXERCISES 
Find  similarly  the  meaning  of : 
1.    4"3.  3.    (f)-2.  5.    a-3.  7.    a)~l 

3 


2.    2-\                  4.  (-5)"6.  6.  a"*.  8.  (-f)  • 
State  the  value  of : 

9.   4"~*.              13.  16_L  17.  32~*.  21.  or3. 

10.   8"*.              14.  16~*.  18.  2T*.  22.  (a0)"2. 


-._?. 


.3 


11.  8~*.  15.  .125"*.  19.  .36  *  23.   a   •. 

12.  (.2°)^  16.  .125"i  20.    64~l  24.    a"5. 


EXPONENTS  65 


WRITTEN     EXERCISES 

Perform  the  operations  indicated  : 

1.    (42-54)~i         .       4.   2"*-2i  7 


„_1      „3 


2.   214-2-6.  5.   2T*-2T*. 


5° 
8.    10-5  •  102  •  10n. 


3.   2*.2~*  6.   3«.3"2.4°.  9.   10r-10-7. 

10.  27"*- 9*.  16.  (a-4)2.         17.    a°-aK 

11.  -^18^.^18^.  is.  i!(0ra-^H)). 

12.  -v'io^4  •  ^lol  a/  ,  , 

2  19.  a2  -a  2(or  a2  2). 

13.  (10-4  •  io-3y.  1 

20.  ^(or«H4)). 
a-2 

15.    102  •  10*  •  10*  •  (A)0.  21.  (<T*)~*         22.    (x-*)s 


14.    10'*  ■  10= 


USE  OF  ZERO,  NEGATIVE,  AND  FRACTIONAL  EXPONENTS 

117.  We  have  defined  zero  and  negative  exponents  so  that 
Law  I  holds  for  them.  It  can  be  shown  that  the  other  four 
laws  hold  for  these  exponents  as  defined,  but  the  laws  will  be 
applied  here  without  proof. 

118.  The  relation  a~n  •  an  =  a0  =  1  can  bem  used  to  change  the 
form  of  expressions. 

I.  To  free  an  expression  from  a  negative  exponent,  multiply 
both  numerator  and  denominator  by  a  factor  that  will  so  combine 
with  the  factor  having  the  negative  exp>onent  as  to  produce  unity 
in  accordance  with  the  relation  just  mentioned.  If  more  than  one 
negative  exponent  is  involved,  apply  the  process  for  each. 

73 .  7-3  •  2*     2* 


For  example  :  7~3  •  24  =  ■ 

i-  1 

h5  .  «3  7-,5~3 

=  b5a3. 


73  73 

Xs       b6  ■  a*      b5az 


b~5     b5  •  6-5        1 

Tkr2  _  fir2  •  ,r~2  _  fi 
r5  ~  fix2  ■  r5  ~  x2 ' 


06  ELEMENTARY   ALGEBRA 


WRITTEN    EXERCISES 

Free  fr< 

Dm 

negative  exponents  : 

o_2 
1.       5-. 

2-3 

4. 

a~lb-2 
x~A 

7. 

2cCK 

10. 

ax~* 

O"5 

2.     —  • 

Zr3 

5. 

a~x 

8. 

ab 
cr*' 

11. 

By3 

Q_2 

3.     3      ' 

5" 

4" 

-3 

i 

6. 

a-3 
bs' 

9. 

a-'b-2. 

12. 

1 

a~3x~3 

II.  To  free  an  expression  from  a  fractional  form,  multiply 
both  numerator  and  denominator  by  a  factor  that,  in  combination 
with  the  denominator,  will  produce  unity.  If  more  than  one  such 
form  is  involved,  apply  the  process  for  each. 

For  example  : 

b*      b-W 

Is  JL        a3         45  ■  2  3r2y8a8 

a&°'  4-5      xV3      45.4-5     x-hfzhj-* 

—  45  . 2  +        x~~y  a' 

(x-'2x--)(2/V3) 

=  2.45  +  a3x-2«/3. 


WRITTEN   EXERCISES 
Free  from  fractional  forms : 


1. 

3 

52' 

3. 

a 

bx3 

5.  i  +  »: 

52     23 

n 

^-1      a;-3 

2. 

1 

a263' 

4. 

a~2 

r3' 

6.    *  +  £• 

y3     Xs 

8.  2+*l 

6      6"1 

EXPONENTS  67 

III.  To  transfer  any  specified  factor  from  the  numerator  into 
the  denominator,  or  vice  versa,  multiply  the  numerator  and  de- 
nominator by  a  factor  that,  in  combination  with  the  factor  to  be 
transferred,  will  produce  unity. 

EXAMPLES 

1.  Transferring  the  factors  of  the  denominator  to  the  numerator  : 

x*y~s     x~iyixiy~3 

2.  Transferring  the  literal  factors  of  the  numerator  to  the  denomi- 
nator : 

5  am  _  5  g-HrWb0-  _      5 
4  a46-3     4  a-:ib-Wb-s     4  ab~5 ' 


WRITTEN     EXERCISES 
In  the  following  expressions : 

(a)  Transfer  all  literal  factors  to  the  numerator. 

(b)  Transfer  all  literal  factors  to  the  denominator. 


1. 

a2x? 

4. 

I,2 

'5  a  *b* 
8ah~i 

7. 

r    -I 
5a  3 

abx 

2. 

6  ay2z~5 
11  ct3x~3y* 

5. 

a  sb-°c* 
ah~5c~* 

8. 

b  x  -\?/- 
5pq 

3 

4  a  W 

17a365c-3' 

6. 

7(cr7b-5)\ 
6  a~M~i 

9. 

3 

a* 

119.  The  laws  of  exponents  enable  us  to  perform  operations 
with  polynomials  containing  fractional  and  negative  ex- 
ponents. 

Thus  :  (a*  +  6^)2  _  (ffi!)2  +  2  ah^  +  (ft*)2 

=  J  +  2  a*6*  +  b. 

OF  THE 


08 


ELEMENTARY   ALGEBRA 


WRITTEN    EXERCISES 
Perform  the  indicated  operations : 


1. 

;o*  -  ity. 

8. 

a5  —  x* 
a^  —  x2 

2. 

:4  oW  +  23)2. 

9. 

(xn-ynf. 

3.    1 

>-*  +  &*)*. 

10. 

(a2nb3r  -  If. 

4.    ( 

a"  -  3  b'f. 

*  4  -  y  3)0 

*  +  y 

■f). 

11. 

ap  •  bq 

5.    1 

av+\  .  ^-i 

6. 

>-"  +  l)(ar» 

-i). 

12. 

(an  +  tn)2. 

7.    1 

[SB*  +  3)(a£  + 

5). 

13. 

(u3&2  +  o:  2)(a362- 

-a;  2). 

Express  as  a  product  of  two  factors 


14.    x6 


m 


-4 


15.    a8-2o4a;'  +  a;l 


16.  a2n  +  2anbn  +  b2n. 

17.  sb*»  —  4  x2"?/2"  +  4  ?/*". 


18.  y~7  —  x~10. 

19.  of  — 4. 

20.  l  +  8x~%  +  16x-5. 

21.  x12  +  6  a?y~*  -f  9  w"7- 


SUMMARY 


I.    Definitions. 


Meaning  of  the  Fractional  Exponents,     a"  denotes  the    qth. 
root  of  the  ^>th  power,  or  the  pth  power  of  the  gth  root  of  a. 

Sec.  107. 

Meaning  of  the  Exponent  Zero.     Any  number  (not  0)   with 
the  exponent  zero  equals  1.  Sec.  115. 

Meaning  of  Negative  Exponents.    a~"  means  —  for  all  values 


a" 


of  n,  positive  or  negative,  integral  or  fractional. 


Sec.  116 


EXPONENTS  69 

U     Laws  of  Exponents. 

1.  am-ar  =  am+r.  Sec.    99. 

2.  am  -=-  «r  =  am"r.  Sec.  100. 

3.  (am)r  =  amr.  Sec.  101. 

4.  (aft)"  =  an6n.  Sec.  101. 

5.  f«Y  =  — •  Sec.  101. 

These  laws  apply  for  all  values  of  the  exponents,  m,  n,  r, 
positive,  negative,  integral,  or  fractional. 

III.    Processes  with  Exponents. 

1.  When  the  bases  of  the  factors  are  the  same,  the  exponent 
of  the  product  is  found  by  adding  the  given  exponents  (Law  I). 

Sec.  109. 

2.  When  the  bases  of  the  expressions  are  the  same,  the 
exponent  of  the  quotient  is  found  by  subtracting  the  exponent 
of  the  divisor  from  the  exponent  of  the  dividend  (Law  II). 

Sec.  110. 

3.  When  an  exponent  is  applied  to  a  number  having  an  ex- 
ponent, the  product  of  the  exponents  is  taken  as  the  exponent 
of  the  result  (Law  III).  Sec.    111. 

4.  An  exponent  affecting  a  product  is  applied  to  each  factor 
(Law  IV).  Sec.  112. 

5.  An  exponent  affecting  a  fraction  is  applied  to  both  nu- 
merator and  denominator  (Law  V).  Sec.  112. 

6.  When  the  bases  are  different  and  the  fractional  exponents 
are  different,  the  exponents  must  have,  or  be  made  to  have,  a 
common  denominator,  before  any  simplification  by  multiplica- 
tion or  division  is  possible.  .  Sec.  113. 

7.  To  free  an  expression  from  a  negative  exponent,  multiply 
both  numerator  and  denominator  by  a  factor  that  will  so  com- 
bine with  the  factor  having  the  negative  exponent  as  to  produce 
unity,  in  accordance  with  the  relation  a-"  •  a"  =  an.       Sec.  118. 


70  ELEMENTARY   ALGEBRA 

8.  To  free  an  expression  from  the  fractional  form,  multiply 
numerator  and  denominator  by  a  factor  that,  in  combination 
with  the  denominator,  will  produce  unity.  If  more  than  one 
such  form  is  involved,  apply  the  process  for  each.         Sec.  118. 

9.  To  transfer  any  specified  factor  from  the  numerator  into 
the  denominator,  or  vice  versa,  multiply  the  numerator  and 
denominator  by  a  factor  that,  in  combination  with  the  factor 
to  be  transferred,  will  produce  unity.  Sec.  118. 

REVIEW 

WRITTEN   EXERCISES 

Express  with  positive  exponents : 

1.  m~*ns.         2.  ±x~%y-lz.         3.  3  a~5bs.         4.  17  aT%-7z~*. 

Transfer  all  literal  factors  from  the  denominator  to  the 
numerator : 

5.     *t.  6.    -J*-.  7.  1      .  8.      5 


x~i'  arh-3  6x~2y*  aV  * 

Multiply : 

9.    (2+V^+T)2.  12.  p.p-K 

10.  V5  .  V&.  13.    (or1  -  b-l)(aT?  -  ft-*). 

11.  5^/mr3  .  2m-1.  14.    (a-2  - 1)(J  -pi). 
Remove  the  parentheses : 


<n-y  21. 

\    _p 

16.    (i/a^)3.  pnu  X  9 

in        \  rfi     )  10n  /       P9\  x2y 

19-     ^    I      •  22.    (»*)*. 

Vlf)  20.     [x^J*1*.  23.     [s~l) 


EXPONENTS  71 


SUPPLEMENTARY   WORK 


ADDITIONAL  EXERCISES 
Perform  the  operations  indicated: 

1 


2. 
3. 

4. 
5. 
6. 

7. 


'  Xo.    zn^   •    —         ' 

16  a*  tor4)*. 
5aafy*)*. 


17.    a;-3  •  x*  •  .t2  •  Vic. 
18. 


Vx     cc3      Vas2 . 


.1.2 


-2B0a^-V»)*  20-    [(«"¥]" 

-243  o-V1^)"*.  21'    lV«ft-*V^{4. 

'"^i^T*.  22-    K^2)*!""1- 


i  i 


8 

9.  Vz-V'V*  23-   ^«2V^ 

io.  v^^.  24-  t>rT°-[(-ocp- 

n  q    H*    ,    o    K*  25-    -a3&-V3cZ5--a-265c4d-5. 

11.  3a263c_1  •  2a465c. 

,  3  ,  26.    a2x  —  3  w6  •  oW. 

12.  10  a&~M  -f-  2  o-W. 

no  m*-2„    2       *.    3  27-   3 *~*5 ** ' 10 **• 

13.  am&  2nc-2  -4-  abnz~3. 

14.  (3  a"2  -5-  6-2)-5.  28.    Va-lv/a3^«^- 


15.    (ar-ffH-a^a*.  29.    Ix"^^-2)"^^1)/)^. 

Simplify : 

30  2*  +  a! 


"^2  31.    ^af-1-*-;*;"*1]2". 


/  cc  4-  2  w 
**~\J>x  +  iy)  32-    [(a^)P"?(«93)?]p: 


3s 


72  ELEMENTARY   ALGEBRA 


(•)  ^I  +  VIo  +  VI-^-    • 

34.  lV45  +  4Vf- V12K 

35.  Express  the  product  Va2  Va3"  as  a  single  radical. 

36.  Divide  2x*y-s-5xiy-2  +  7x*y-l-5x%  +  2xiy  by 

Find  equivalent  expressions  with  rational  divisors : 

37.   3V2^2V36.  48    (V5-2)(3+V5), 

5-Vo 

p-  Vg 


38.  &V(?-^Va6. 

39.  V40  ic3?/  -5-  a;  V5  ?/.  49- 


p+  Vg. 
40.   .-^y  +  sf^  z+V^l 


50. 


41.  2^2a2-\/4a.     '  z-V^T 

42.  _|Vf^TVV3.  51  V5 

Va  —  V& 


43.  6A/54.r2-2A/'2x2. 

44.  a2^48^-r-2a&^3aP:  52.    Vs-Vs  +  y, 

Va;  +  Va;  +  2/ 

45.  4  OOJ  -5-  V('.*'.  

/ „     Va24-/>2  +  & 

46.  V«  +  ft  +  YJLzJ.  M"    Va^+F-& 
Va+6— Va— & 


47. 


1  **     3Va-3  +  V.r-f  3 

54. 


a  _  Va2  -  x2  3  Va;  —  3  —  Va;  +  3 


EXPONENTS  73 


V^+T+V^T                      J3  +  V5-V5-V5 
.  fifi'  j • 


55.  ' —  •  56 


1 


57. 


2  +  Vo  -  V2 


Suggestion.    Rationalize  the  denominator  in  two  steps,  using  as  first 

factor  ,-         _ 

2_(\/5-  V2). 

Solve :  - 

58.    3*+ Va*-2a;  +  5  =  l.       59-   5 =  a2 -±  2  ab  +  b%' 


^_ _i_  :  /,,.  .  ,„  .   ...  ..  .  ,   ,    2a& 

+  - 


a  +  0  a+6 


61.   Vz(2a-&+ Va)  =  3a2  — a&. 


CHAPTER    V 

LOGARITHMS 

MEANING  AND  USE  OF  LOGARITHMS 

120.  Use  of  Exponents  in  Computation.  By  applying  the 
laws  of  exponents  certain  mathematical  operations  may  be 
performed  by  means  of  simpler  ones.  The  following  table  of 
powers  of  2  may  be  used  in  illustrating  some  of  these  simpli- 
fications : 


1  =2° 

32  =  25 

1024  =  2io 

32768  =  215 

2  =  2i 

64  =  26 

2048  =  2U 

65536  =  2!6 

4  =  22 

128  =  27 

4096  =  2i2 

131072  =  2" 

8  =  23 

256  =  28 

8192  =  213 

262144  =  2" 

16  =  2* 

512  =  29 

16384  =  2" 

524288  =  219 

121.    Application  of  Law  I,  Sec.  99,  p.  53. 

EXAMPLES 

1.  Find:    8-32. 

From  the  table,  8  =  23,  (1) 

and  32  =  25.  (g) 

Then,  8  •  32  =  23  •  25  =  28,  (J) 

and,  according  to  the  table,  28  =  256.  (^) 

2.  Find:    2048-64. 

From  the  table,                                        2048  =  211,  (i) 

and                                                     64  =  2G.  (2) 

Then,  2048  •  64  =  2ii  .  26  _  2",  (3) 

and,  according  to  the  table,                 217  =  131072.  (4) 

Thus  the  process  is  simply  one  of  inspection.  In  the  above  example 
we  merely  added  11  and  6  and  looked  in  the  table  for  the  number 
opposite  to  217. 

74 


LOGARITHMS  75 

ORAL   EXERCISES 
State  the  following  products  by  reference  to  the  table : 

1.  16  •  256.  5.   32  •  32.  9.    128  •  512. 

2.  32-128.  6.   64-64..  10.    128-1024. 

3.  64-512.  7.   32-2048.  11.   8-16384. 

4.  8  -  2048.  8.   16  •  4096.  12.   32  -  4096. 

122.    Application  of  Law  II.  Sec.  100,  p.  54. 

EXAMPLES 


(3) 
(4) 


1.  Find:.  256. 
32 

From  the  table, 

256  =  28, 

and 

32  =  25. 

Hence, 

256 
32 

98 

=  —  =  28-5=28, 

25                               ' 

and,  according1 

to  the  table, 

23=8. 

2.  Find:    65536. 
2048 

As  above, 

65536 
2048 

016 

=  —  =  25  =  32. 

2ii 

ORAL   EXERCISES 
By  use  of  the  table  determine  the  value  of  the  following : 
,      1024  „     32768  32  •  2048 

1.     •  6.     •  5. • 

128  1024  512 

2     8192  64  -  512  128-  131072 

64   '  '    16-128'  '     64-8-8192  ' 

123.    Application  of  Law  III,  Sec.  101,  p.  54. 

EXAMPLES 

1.    Find:    163. 

By  the  table,  16  =  24.  (1) 

Hence,  163=  (24)3  =  212.  (£) 

and,  according  to  the  table,     21'2  =  4096.  (5) 


7.    V8192. 

10. 

11. 
12. 

5121. 

8.    ^4096. 

a/32768. 

9.    ^65536. 

a/1024. 

76  ELEMENTARY   ALGEBRA 

2.  Find:   a/1024. 

As  above,  1024  =  (1024)*  =  (21*)*  _  25  =  32,  according  to  the  table. 

3.  Find :  a/32768. 

As  above,  \^2768  =  (215)  £  _  23  =  8. 

ORAL  EXERCISES 
By  use  of  the  table  find  the  value  of : 

1.  323.  4.   642. 

2.  35.  5.   2562 

3.  325.  6.   164. 

124.  The  examples  and  exercises  above  show  that  the  laws 
of  exponents  furnish  a  powerful  and  remarkably  easy  way  of 
making  certain  computations. 

In  the  above  illustrations  we  have  used  a  table  based  on  the  number  2, 
and  have  limited  the  table  to  integral  exponents  ;  but  for  practical  pur- 
poses a  table  based  on  10  is  used  and  is  made  to  include  fractional 
exponents. 

For  example : 

1.  It  is  known  that  approximately, 

2  =  lO1^  or  10-3  (more  accurately  10-301). 
From  this  we  can  express  20  as  a  power  of  10,  for 

20  =  10  •  2  =  101  •  10-301  =  101-301. 
Similarly,  200  =  10  •  20  =  101  •  lO1*"  =  102-301, 

and  2000  =  10  •  200  =  101  •  lO***  =  103-301. 

2.  It  is  known  that  approximately  763  =  102-88. 
Then 
and 

Similarly, 
and 


7630  = 

=  10- 

763=  10i -102-88  =103-88, 

6300  = 

=  100 

.  763  =  102  •  102-88  =  10^-88. 

76.3  = 

_  763 
10 

1<V  88 
_  *u        _  JQ2.88-1  —  101-88 

101 

7.63  = 

763 
"100" 

1fl2  88 
_  1"        _  1Q2.88-2—  100-88 

102 

LOGARITHMS  J  77 

WRITTEN   EXERCISES 

Given  48  =  10168 ;  express  as  a  power  of  10 : 
1.  480.  2.  4800.  3.  48,000.  4.  4.8. 

Given  649  =  102-81;  express  as  a  power  of  10 : 

5.  6490.  7.  649,000.  9.  6.49. 

6.  64,900.  8.  64.9.  10.  649,000,000. 

Given  300  =  10247;     express  as  a  power  of  10 : 

11.  3.  13.  3000.  15.  300,000. 

12.  30.  14.  30,000.  16.  3,000,000. 

125.   The  use  of  the  base  10  has  several  advantages. 

I.  The  exponents  of  numbers  not  in  the  table  can  readily  be 
found  by  means  of  the  table. 

To  make  this  clear,  let  us  suppose  that  a  certain  table  expresses  all 
integers  from  100  to  999  as  powers  of  10  ;  then  80,  although  not  in  this 
table,  can  be  expressed  as  a  power  of  10  by  reference  to  the  table. 

For,  30  =  ^- ,  and  since  300  is  in  the  supposed  table  we  may  find 
10 

1  02-47 

by  reference  to  the  table  that  300  =  102-*7,  and  hence,  30  =  ^—  =  101-47. 
J  101 

Similarly,    3.76    is    not    in    the    supposed    table,    but    376    is    and 

3.76  =  -—  =  -— •       Therefore  it  is  necessary  only  to  subtract  2  from 
100      102 

the  power  of  10  found  for  376  in  order  to  find  the  power  of  10  equal  to 

3.76. 

Similarly,  4680  is  not  in  the  table,  but  468  is  and  4680  =  408  ■  101. 
Therefore,  it  is  necessary  only  to  add  1  to  the  power  of  10  found  for  468 
in  order  to  find  the  power  of  10  equal  to  4680. 

Such  a  table  would  not  enable  us  to  express  in  powers  of  10  numbers 
like  4683,  46.83,  and  356,900,  but  only  numbers  of  3  or  fewer  digits,  which 
may  be  followed  by  any  number  of  zeros. 

Similar  conditions  would  apply  to  a  table  of  powers  for  numbers 
from  1000  to  9999,  from  10,000  to  99,999,  and  so  on. 

II.  The  integral  part  of  the  exponent  can  be  written  with- 
out reference  to  a  table. 


78  *  ELEMENTARY   ALGEBRA 

For  example  : 

1.  879  is  greater  than  100,  which  is  the  second  power  of  10,  and  less 
than  1000,  or  the  third  power  of  10.  That  is,  879  is  greater  than  10'2  but 
less  than  103.  Therefore  the  exponent  of  the  power  of  10  which  equals 
879  is  2.  +  a  decimal. 

2.  Similarly,  87.9  lies  between  10  and  100,  or  between  101  and  10'3, 
hence  the  exponent  of  the  power  of  10  that  is  equal  to  87.9  is  1.  +  a 
decimal. 

ORAL    EXERCISES 

State  the  integral  part  of  the  exponent  of  the  power  of  10 
equal  to  each  of  the  following : 

l.  35.  4.  25.  7.  25.5. 

2    350.  5.  2500.  8.  3G5.5. 

3.  36.5.  6.  36,500.  9.  17.65. 

III.  If  two  numbers  have  the  same  sequence  of  digits  but 
differ  in  the  position  of  the  decimal  point,  the  exponents  of 
the  powers  of  10  which  they  equal  have  the  same  decimal 
part. 

For  example : 

Given  that  274.3  =  102-43, 

we  have  27.43  =  ^M  =  ™^  =  101 « 

10  10i 

also  2743  =  10  •  274.3  =  101  •  102-43  =  103-43, 

also  274,300  =  1000 •  274.3  =  103  •  102*3  =  105-43. 

In  each  instance  the  decimal  part  of  the  exponent  is  the  same.  It  is 
evident  that  this  will  be  the  case  in  all  similar  instances,  for  shifting  the 
decimal  point  is  equivalent  to  multiplying  or  dividing  repeatedly  by  10, 
which  is  equivalent  to  changing  the  integral  part  of  the  exponent  by 
adding  or  subtracting  an  integer. 

ORAL    EXERCISES 

Given  647  =  10281,  state  the  decimal  part  of  the  exponent 
of  the  power  of  10  that  equals  : 

1.  64.7.  3.   6470.  5.   647,000. 

2.  6.47.  4.    64,700.  6.   6,470,000. 


LOGARITHMS  79 

Given  568.1  =  102-75,  state  the  decimal  part  of  the  exponent 
of  the  power  of  10  that  equals  : 

7.  56.81.  9.   5681.  11.   568,100. 

8.  5.681.  10.   56,810.  12.  56,810,000. 

126.  Logarithms.  Exponents  when  used  in  this  way  for 
computation  are  called  logarithms,  abbreviated  log. 

127.  The  number  to  which  the  exponents  are  applied  is 
called  the  base. 

For  the  purposes  of  computation  the  base  used  is  10. 

According  to  the  above  definition  the  equation  30  =  10 lAS  may  be 
written  log  30  =  1.48,  which  is  read  "  The  logarithm  of  30  is  1.48."  These 
equations  mean  the  same  thing  ;  namely,  that  1.48  is  (approximately) 
the  power  of  10  that  equals  30. 

WRITTEN    EXERCISES 
Write  the  following  in  the  notation  of  logarithms  : 

1.  700  =  10284.  3.   6  =  10077.  5.   361  =  10255. 

2.  75  =  10187.  4.   50  =  10169.  6.   45  =  10165. 

Write  the  following  as  powers  of  10 : 

7.  log  20  =  1.3.  9.   log  3  =  0.47.        11.   log  111  =  2.04. 

8.  log  500  =  2.70.        10.    log  7  =  0.84.        12.    log  21  =  1.32. 

128.  Characteristic  and  Mantissa.  The  integral  part  of  a 
logarithm  is  called  its  characteristic,  and  the  decimal  part  its 
mantissa. 

ORAL    EXERCISES 

1-12.  State  the  characteristic  and  the  mantissa  in  each  of 
the  logarithms  in  Exercises  1-12  above. 

129.  According  to  Sec.  125,  II  (p.  77),  the  characteristics 
of  logarithms  can  be  determined  by  inspection,  consequently 
tables  of  logarithms  furnish  only  the  mantissas. 


80  ELEMENTARY   ALGEBRA 


EXPLANATION  OF  THE   TABLES 

130.  The  use  of  the  tables,  pp.  84  and  85,  is  best  seen  from 
an  example. 

Find  the  logarithm  of  365. 

The  first  column  in' the  table,  p.  84,  contains  the  first  two  figures  of 
the  numbers  whose  mantissas  are  given  in  the  table,  and  the  top  row  con- 
tains the  third  figure. 

Hence,  find  36  in  the  left-hand  column,  p.  84,  and  5  at  the  top. 

In  the  column  under  5  and  opposite  to  36  we  find  5623,  the  required 
mantissa. 

Since  365  is  greater  than  100  (or  102)  but  less  than  1000  (or  103),  the 
characteristic  of  the  logarithm  is  2. 

Therefore,  log  365  is  2.5623. 

WRITTEN    EXERCISES 
By  use  of  the  table  find  the  logarithms  of  : 

1.  25.  5.  99.  9.  9.9.  13.  1000. 

2.  36.  6.  86.  10.  8.6.  14.  5000. 

3.  50.  7.  999.  11.  33,000.  15.  505. 

4.  75.  8.  800.  12.  99,900.  16.  5.05. 

131.  Negative  Characteristics.  An  example  will  serve  to 
show  how  negative  characteristics  arise : 

From  log  346  =  2.5391,  we  find, 

log 34.6  =  log  —    =  log 346  -  log  10  =  2.5391  -  1  =  1.5391. 

log  3.46  =  log  —  =  1.5391  -  1  =  0.5391. 

log  .346  =  log  —  =  0.5391  -  1. 
&  &   10 

In  the  last  line  we  have  a  positive  decimal  less  1,  and  the  result  is  a 
negative  decimal ;  viz.  —  .4609.  But  to  avoid  this  change  of  mantissa,  it 
is  customary  not  to  carry  out  the  subtraction,  but  simply  to  indicate  it. 
It  might  be  written  —  1  +  .5391,  but  it  is  customarily  abridged  into 
1.5391.  The  mantissa  is  kept  positive  in  all  logarithms.  The  loga- 
rithm 1.5391  says  that  the  corresponding  number  is  greater  than  10 -1 
(or  Jj),  but  less  than  10°  or  1. 


LOGARITHMS  81 


We  now  write  log   .346  =  1.5391 


» 


Similarly,  log  .0346  =  log  —  =  1.5391  -  1  =  2.5391. 

Thus  we  see  that  the  mantissa  remains  the  same,  no  matter  how  the 
position  of  the  decimal  point  is  changed.  The  mantissa  is  determined 
solely  by  the  sequence  of  digits  constituting  the  number. 

The  characteristic  is  determined  solely  by  the  position  of  the  decimal 
point.  Xegative  characteristics,  like  positive  ones,  are  determined  by 
inspection. 

EXAMPLES 

1.  What  is  the  characteristic  of  log  .243  ? 

.243  is  more  than  .1  or  10-1,  but  less  than  1  or  10°.     Hence, 

.243  =  10_1+  a  decimal. 

The  characteristic  is  1. 

2.  Find  the  characteristic  of  log  .0593. 

.0593  is  more  than  .01  or  10~2,  but  less  than  .1  or  10_1.     Hence, 

.0593  =  10-2+ a  decimal. 

The  characteristic  is  —  2,  or  2. 

3.  Similarly,  since  .00093  is  greater  than  .0001  or  10  ~4,  but 
less  than  .001  or  10~s, 

log  .00093  =  4+  a  decimal. 

132.  The  characteristic  having  been  determined,  the  mantissa 

is  found  from  the  table  in  the  usual  way. 

For  example  : 

log      .243  =  1.3856, 

log    .0593  =  2.7731, 

log  .00093  =  4.9685. 

WRITTEN    EXERCISES 

Find  the  logarithms  of : 

1.  .35.  3.   .105.  5.   .0023.  7.   .00342. 

2.  ,j.634.  4.    .027.  6.    .0123.  8.    .0004. 

133.  In  finding  the  number  corresponding  to  a  logarithm 
with  negative  characteristic,  the  same  method  is  followed  as 
when  the  characteristic  is  positive.  The  mantissa  determines 
the  sequence  of  digits  constituting  the  number ;  the  character- 
istic fixes  the  position  of  the  decimal  point. 


32 


ELEMEXTARY   ALGEBRA 


For  example  : 

If  log  n  —  2.595o.  the  digits  of  n  are  394.     The  characteristic  2  says 
that  n  is  greater  than  10--  out  less  than  10-1  (or  .1).     Hence. 

the   decimal   point   must   be   so  placed  that  n  has  no  tenths  but  some 
hundredths.     Therefore  n  =  .0394. 


WRITTEN    EXERCISES 
By  use  of  the  tables  find  the  numbers  whose  logarithms  are 


1.  1.6232. 

2.  1.4914. 

3.  2.4281. 

4.  1.9196. 

5.  0.9196. 

6.  3.4281. 

Find  n  if : 

19.  log  n  =  1.9289. 

20.  log  n=  0.9289. 


7.  2.0792 

8.  -.7076 

9.  4.9196 

10.  3.2201 

11.  0.2201 

12.  1.2201 


13.  3.0969. 

14.  1.6972. 

15.  3.9284. 

16.  2.9284. 

17.  1.9284. 

18.  5.7832. 

21.  log  (n  — 1)  =  3.9294. 

22.  log  (|  n)  =  1.6128. 


I.      10s*  •  10r  =  10"^r. 


USE   OF   THE    TABLES   FOR   COMPUTATION 

134.   For  use   in  computation   by  logarithms   the    laws    of 
exponents  may  be  expressed  thus  : 

Tfte   logarithm  of  a   product  is  the 

sum  of  the  logarithms  of  the  factors. 

Thie  logarithm  of  a  quotient  is  the 
logarithm  of  the  dividend  minus  the 
logarithm  of  the  divisor. 

The  logarithm  of  a  number  with  an 
exponent  is  the  product  of  the  exponent 
and  the  logarithm  of  the  number. 


II. 


10"* 

—  =1'    . 

irr 


III.     (10-)'  =  10" 


Since  r  may  be  positive  or  negative,  integral  or  fractional,  Law  III 
provides  not  only  for  raising  to  integral  powers,  but  also  for  finding  recip- 
rocals of  such  powers,  and  for  extracting  roots. 


LOGARITHMS  83 

EXAMPLES 

1.  Multiply  21  by  37. 

1.  log  21  =  1.3222,  table,  p.  84. 

2.  log 37  =  1.5682,  table,  p  84. 

3.  Adding,  log 21  +  log 37,  or  log  (21  x  37)=  2.8904  (Sec.  134). 

4.  .•.  21  x  37  =  777  from  table,  p.  85. 

2.  Divide  814  by  37. 

1.  log 814  =  2.9106,  table,  p.  85. 

2.  log  37  =  1.5682,  table,  p.  84. 

3.  .-.  log  814  -  log 37  =  2.9106  -  1.5682  =  1.3424. 

4.  .-.  814  -h  37  =  22,  table,  p.  84. 

3.  Extract  the  cube  root  of  729. 

1.  ^729  =  (729)i 

2.  log  7293  _  i  log  729  (Sec.  134). 

3.  log  729  =  2.8627,  table,  p.  85. 

4.  i  of  2.8627  =  0.9542. 

5.  .-.  (729)3  or  .3/729  _  g?  table,  p.  85. 

Compute  by  use  of  logarithms: 

1.  8  x  15.  6.  54. 

2.  41  x  23.  7.  312. 

3.  37  x  17.  8.  V414. 

4.  12x17.  9.  V343. 

5.  893-5-19.  10.  940  h- 47.              15.    V216 

Since  the  logarithm  is  approximate,  the  result  in  general  is  approxi- 
mate. Thus,  log  \/(256)  =  0.60205,  which  is  not  the  logarithm  of  4,  but 
is  sufficiently  near  to  be  recognized  in  the  table. 


11. 

192. 

12. 

73. 

13. 

45. 

14. 

V196. 

1  c 

3  /„ .  n 

84 


ELEMENTARY   ALGEBRA 


N 
10 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

11 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1106 

13 

1139 

1173 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

14 

1461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

1703 

1732 

15 

17G1 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

18 

2553 

2577 

2601 

2625 

2(i48 

2672 

2695 

2718 

2742 

2765 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

23 

3617 

3636 

3655 

3674 

3692 

3711 

3729 

3747 

3766 

3784 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

5038 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145- 

5159 

5172 

33 

5185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

36 

5563 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

5670 

37 

5682 

5694 

5705 

5717 

571".) 

5740 

5752 

5763 

5775 

5786 

38 

5798 

5809 

5821 

5832 

5843 

5855 

-,S66 

5877 

5888 

5899 

39 

5911 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

41 

6128 

613S 

DIVJ 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6:  ;■_'.", 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

45 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

46 

6628 

6637 

6646 

6056 

666,5 

6675 

6684 

6693 

6702 

6712 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

48 

6812 

6821 

6830 

6S39 

6848 

6.S57 

6S66 

6875 

6SS4 

6893 

49 

6902 

6911 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

50 

6990 

6998 

7007 

7016 

7021 

7033 

7042 

7050 

7059 

7067 

51 

7076 

7084 

7093 

7101 

7110 

7118 

7126 

71 35 

7143 

7152 

52 

7160 

7168 

7177 

71S5 

7193 

7202 

7210 

7218 

722(5 

7235 

53 

7243 

7251 

72:,!  I 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

LOGARITHMS 


85 


N 
55 

0 

1 

2 

3 

1 

4 

5 

6 

7 

8 

9 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

56 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

58 

7634 

7642 

7649 

7657 

7664 

7672' 

7679 

7686 

7694 

7701 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

61 

7853 

7860 

7868 

7875 

7-S.S2 

7889 

7896 

7903 

7910 

7917 

62 

7924 

7931 

793S 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

71 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

72 

8573 

8579 

S5S5 

8591 

8597 

8603 

8609 

8615 

8621 

8627 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

77 

ssi;.-. 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

86 

9345 

9350 

9355 

9:360 

9365 

9370 

9375 

9380 

9385 

9390 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

88 

9445 

9450 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

89 

9494 

9499 

9504 

9509 

9513 

9518 

9523 

9528 

9533 

9538 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

91 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

93 

9685 

9689 

!Ki94 

9699 

9703 

970S 

9713 

9717 

9722 

9727 

94 

9731 

9736 

9741 

9745 

9750 

9754 

9759 

9763 

9768 

9773 

95 

9777 

97S2 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

96 

9823 

9827 

9832 

9836 

9841 

9S45 

9850 

9854 

9859 

9863 

97 

9S68 

9872 

9877 

9881 

9886 

9890. 

9894 

9899 

9903 

9908 

98 

9912 

9917 

9921 

9926 

9930 

9934 

9939 

994:'. 

9948 

9952 

99 

N 

9956 

9961 

9965 

9969 

9974 

9978 

99S3 

9987 

9991 

9996 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

86  ELEMENTARY  ALGEBRA 

135.    How  to  calculate  the  logarithm  of  a  number  not  found 
in  the  table  is  best  seen  from  an  example. 

EXAMPLE 
Find  the  logarithm  of  257.3  (see  16th  line  of  table,  p.  84) : 


0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

25 

3979 

3997 

4014 

4031 

4048 

4i  Mi:. 

4082 

4099 

4116 

4133 

The  numbers  nearest  to  257.3  whose  logarithms  are  given  in  the  table 
are  257  and  258.     We  have 

log  258  =  2.4116 

log  257  =  2.4099 

Difference,     .0017 

That  is,  an  increase  of  1  in  the  number  causes  an  increase  of*  .0017  in  the 
logarithm.  Assuming  that  an  increase  of  .3  in  the  number  would  cause 
an  increase  of  .3  of  .0017,  or  .0005,  in  the  logarithm,  then 

log  257.3  =  2.4099  +  .0005  =  2.4104. 

Notes  :  1.  The  product  of  .3  and  .0017  is  .00051,  but  we  take  only  four 
places,  because  the  mantissas  as  given  in  the  table  are  expressed  to  four 
places  only.  If  the  digit  in  the  fifth  decimal  i>lace  of  the  correction  is 
more  than  5,  we  replace  it  by  a  unit  in  the  fourth  place. 

2.  The  difference  between  two  succeeding  mantissas  of  the  table 
(called  the  tabular  difference)  can  be  seen  by  inspection. 

3.  What  is  written  in  finding  the  logarithm  of  257.3  should  be  at  most 
the  following : 

log  257  =  2.4099  tabular  difference    17 

correction  for  .3  = 5  .3 

log  257.3  =  2.4104  IT 

4.  The  corrections  are  made  on  the  assumption  that  the  change  in  the 
logarithm  is  proportional  to  the  change  in  the  number.  This  is  suffi- 
ciently accurate  when  used  within  the  narrow  limits  here  prescribed. 

WRITTEN    EXERCISES 
Find  the  logarithm  of : 

1.  1235.        5.   1425.          9.   3.142.       13.  .4071.  17.  .3002. 

2.  23.5.        6.   1837.        10.   1.414.       14.  85.51.  18:  9009. 

3.  2.36.         7.   6720.        11.    1.732.       15.  .OUT..  19.  12.02. 

4.  .0237.       8.    67.25.        12.    .6220.        16.  .ll'67.  20.  5.008, 


LOGARITHMS 


81 


136.  To  calculate  the  number  whose  logarithm  is  given 
apply  the  table  as  follows: 

(1)  If  the  given  logarithm  is  in  the  table,  the  number  can 
be  seen  at  once. 

(2)  If  the  given  logarithm  is  not  in  the  table,  the  number 
corresponding  to  the  nearest  logarithm  of  the  table  may  be 
taken. 

A  somewhat  closer  approximation  may  be  found  by  using 
the  method  of  the  following  example  : 


EXAMPLE 


Find  the  number  whose  logarithm  is  1.4271.     The  mantissas 
nearest  to  this  are  found  in  the  17th  line  of  table,  p.  84. 


26 


0 


4150  410G  4183  4200  4210 


8 


4232     4249     4265      4281      4298 


The  process  is  the  reverse  of  that  of  finding  the  logarithm. 

The  next  smaller  mantissa  in  the  table  is  4205,  corresponding  to  the 
number  267.  The  difference  between  this  mantissa  and  the  given  man- 
tissa is  6.  The  tabular  difference  between  4205  and  the  next  larger  man- 
tissa is  10.  An  increase  of  10  in  the  logarithm  corresponds  to  an  increase 
of  1  in  the  number.  Hence,  an  increase  of  0  in  the  logarithm  corresponds 
to  an  increase  of  TR5  of  1,  or  .4,  in  the  number.  This  means  .4  of  one  unit 
in  the  number  207.  What  its  place  value  is  in  the  final  result  depends 
upon  the  characteristic.     The  digits  of  the  result  are  2674. 

The  characteristic  1  shows  that  the  desired  number  is  greater  than  the 
first  power  of  10,  but  less  than  the  second  power  of  10  or  100.  Hence, 
the  decimal  point  must  be  placed  between  0  and  7,  and  the  final  result  is 
26.74. 

Notes:  1.  For  reasons  similar  to  those  of  Note  1,  p.  86,  the  correction 
should  be  carried  to  one  place  only. 

2.   At  most  the  following  should  be  written  : 

logn  =  1.4271  tab.  diff.  16 

mantissa  for  207  =    .4265  T6e  =  A 


diff. 


Therefore, 


0 
n  =  26.74. 


88  ELEMENTARY   ALGEBRA 

WRITTEN    EXERCISES 
Find  the  number  whose  logarithm  is : 

1.  0.7305.  4.   2.9023.  7.   1.1962.  10.   3.9485. 

2.  0.5029.  5.   3.1467.  8.   2.0342.  11.   4.6987. 

3.  1.4682.  6.   3.6020.  9.  3.3920.  12.  2.6376. 

SUMMARY 

I.  Definitions. 

1.  Exponents  indicating  the  powers  of  a  base  and  used  for 
the  purposes  of  calculation  are  called  logarithms.  Sec.  126. 

2.  The  integral  part  of  a  logarithm  is  called  its  characteristic, 
and  the  decimal  part  its  mantissa.  Sec.  128. 

II.  Laws. 

1.  10m  •  10r  =  10m+r.  The  logarithm  of  a  product  is  the  sum 
of  the  logarithms  of  the  factors. 

10m 

2.  —  =  10m_r.    The  logarithm  of  a  quotient  is  the  logarithm 

of  the  dividend  minus  the  logarithm  of  the  divisor. 

3.  (10m)r  =  10mr.  The  logarithm  of  a  number  with  an  ex- 
ponent is  the  product  of  the  exponent  and  the  logarithm  of 
the  number.  Sec.  134. 

REVIEW 

WRITTEN     EXERCISES 

Find  exactly  or  approximately  by  use  of  logarithms  the 
value  of: 


1.  V2.  4.    V7.  7.    V756.  10.    (1.03)7. 

2.  V5.  5.    V92U  8.    ^812.  11.    (1.04)10. 

3.  i/d.  6.    a/656.  9.    (1.5)4.  12.    (1.06)9. 


13. 
14. 


M?§).  15.    V(624)(598)(178). 


779 


(732X774),  16     J(651)(654)(558) 

(731)  (671)  '     *  763 


LOGARITHMS  89 

17.  It  is  known  that  the  volume  of  a  sphere  is  §  Ttr3,  r  being 
the  length  of  the  radius.  Using  3.14  as  the  approximate 
value  of  ir,  find  by  logarithms  the  volume  of  a  sphere  of  radius 
7.3  in. 

18.  Find,  as  above,  the  volume  of  a  sphere  whose  radius  is 
36.4  ft. 

Calculate  by  logarithms : 

(132)(1837)  6.    ^^578. 

167 


(2076)  (379) 
173 


(15.61)2 
(700)3 


(3059)(349)  /(7688)(7719) 

(19)(23)(2443)  *    \  (248)(249)  " 

J(294)(1842)>  9    (217.6)(.00681). 


307 

3X675)  1Q      [V278.2  (2.578)1* , 


113  V.00231  •  V76.19 

11.  Given  a  =  0.4916,  c  =  0.7544,  and  b  =  c2  -  a2.     Find  &. 
Suggestion.  6  =  (c  —  a)(c  +  a). 

12.  It  is  known  that  in  steam  engines,  the  piston  head's 
average  velocity  (c)  per  second  is  approximately  given  by  the 
formula : 

31 


= «>w& 


15' 

where  s  denotes  the  distance  over  which  the  piston  moves 
(expressed  in  the  same  unit  as  c),  and  p  the  number  of  pounds 
pressure  in  the  cylinder. 

(1)  Find  c,  if  s  =  32.5  in.,  p  =  110  lb. 

(2)  Findp,  if  c  =  15  ft.,  s  =  2.6  ft. 


CHAPTER   VI 
IMAGINARY  AND   COMPLEX  NUMBERS 

137.  Imaginary  Numbers.  The  numbers  denned  in  what  pre- 
cedes have  all  had  positive  squares.  Consequently,  among 
them  the  equation  ar  =  —  3,  which  asks,  "  What  is  the  number 
whose  square  is  —  3  ?  "  has  no  solution. 

A  solution  is  provided  by  defining  a  new  number,  V—  3,  as 
a  number  whose  square  is  —3.  Similarly  we  define  V  — a, 
where  a  denotes  a  positive  number,  as  a  number  whose  square 
is  —a. 

The  square  roots  of  negative  numbers  are  called  imaginary 
numbers. 

138.  If  a  is  positive,  V  —  a  may  be  expressed  Va  V  —  I. 
Similarly,  V—  5  =  Vo  (—  1)  =  Vo  V—  1. 


V-49  =  V49(-  1)  =  7  V^T. 

139.   Real  Numbers.      In   distinction  from   imaginary  num- 
bers, the  numbers  hitherto  studied  are  called  real  numbers. 

WRITTEN    EXERCISES 
Express  as  in  Sec.  138 : 

1.  V^9.       4.    V^IOO.        7.    V^18.  10.    a/^12. 

2.  V-16.      5.     -V^64.       8.     -V-32.       11.     V-50. 


3.    V-25.      6.    V^8.  9.    -V— 7.         12.    —V- 


i  •>. 


140.    The  positive  square  root  of  —  1  is  frequently  denoted 
by  the  symbol  i ;  that  is,  V—  1  =  i. 


IMAGINARY   AND   COMPLEX   NUMBERS  91 

Using  this  we  write  : 

V^5  =  V5  •  i. 


V^-49  =  ±li. 


V-  75a26  =  V3  . 25 a26' .-1=5  aV3&  •  i. 

Note.    Throughout  this  chapter  the  radical  sign  is  taken  to  mean  the 
positive  root  only. 

WRITTEN    EXERCISES 

Rewrite  the  following,  using  the  symbol  i  as  in  Sec.  140 : 

9.    12  -V-~9. 


1. 

2  +  V-4. 

5.   25-V-25. 

2. 

3-V-9. 

6.    5— V— 3. 

3. 

4+V-4. 

7.    3  +  V-6. 

4. 

5_V-16. 

8.    7  +  V-12. 

13.     -V-62c. 

15. 

10.     2V-100. 


11.  4  V— (a  +  6). 

12.  Va  +  V— 62c2. 


.'• 


+  #—  V—  xy* 


14.   a  +  V-  (a*  +  x2).  16.  p2  +  V-  (p  -h  q)3. 

141.  Complex  Numbers.  A  binomial  one  of  whose  terms  is 
real  and  the  other  imaginary  is  called  a  complex  number. 

The  general  form  of  a  complex  number  is  a  +  bi,  where  a 
and  b  may  be  any  real  numbers. 

Note.  Complex  numbers  are  also  simply  called  imaginary,  any  ex- 
pression which  involves  i  being  called  imaginary.  Single  terms  in  which 
i  is  a  factor  (those  which  we  have  called  imaginary  above)  are  often 
called  pure  imaginaries,  while  the  others  are  called  complex  imaginaries. 
Thus,  V—  2,  3V—  a,  5  i  are  pure  imaginaries  and  1  —  V—  3,  a  —  V—  b 
are  complex  imaginaries. 

ORAL    EXERCISES 

1.  Name  the  real  term  and  the  imaginary  term  in  each 
exercise  of  the  last  set. 

2.  Name  the  values  of  a  and  b  in  each  exercise. 

142  Processes  with  Imaginary  and  Complex  Numbers.  After 
introducing  the  symbol  i  for  the  imaginary  unit  V—  1,  the 
operations  with  imaginary  and  complex  numbers  are  performed 
like  the  operations  with  real  numbers. 


92  ELEMENTARY   ALGEBRA 

I.    Addition  and  Subtraction. 

EXAMPLE 


AddV-9,    -V^25,   V^3. 

V^9  =  3  i. 


■  ( 


\ 


-  V-  25  =  -  5  i. 
\/="3  =  -  V3.i. 
.-.  the  sum  is  (3  -  5  -  \/S)i  =  -  (2  +  V3)i. 

WRITTEN     EXERCISES 
Add: 

1.   2  i,  3  i,  -i.  6.  3  +  4  i,  2  -  3  i,  5  +  5  i. 


2.    Vl6i,  -2i.  7.  V-9z2,  -V-8«2. 


3.  V=16,  -2V:rl.  8.  V-(a  +  &)2,-V(6+c)2. 

4.  V^4,  V^,  a/^1.  9.  2V-32a3,  3V^8a3,  &V2L 

II.    Multiplication. 


To  multiply  complex  numbers  we  apply  the  fact  that  V—  1 
.  V^L  =  —  1,  or  ?2  =  —  1,  since  the  square  of  the  square  root 
of  a  number  is  the  number  itself. 

EXAMPLES 

Multiply  : 

1.    V^l6  by  V:r9. 

V^16  =  4V^l  =  4i\ 


■v/^9    =3V-  1  =Si. 


.:  the  product  is  12(\/^T)'2  =  (12) (-  1)  =  -  12. 
This  may  be  written  (4  i)(3  i)  =  12 12  =  -  12. 

2.     3  -  V^  by  2  -  V^            3-  a  +  bibya-bi. 

S-VSi  a  +  bi 

2  - V5 i  a ~ M 

6-2V3i  a2  +  aM 

-  3V5  i  +  a/1512  -abi-  b2i? 

6-(2\/3  +  3V5)f-  \/l5.  "2            + 6* 


IMAGINARY  AND  COMPLEX  NUMBERS  93 

143.  a  +  bi  and  a  —  bi  are  called  conjugate  complex  numbers. 

WRITTEN    EXERCISES 
Multiply  : 

1.  5-3tby5  +  3i.  7.  4  +  ?-by5  —  i. 

2.  3  +  V^3  by  2  +  V^5.  8.  a  +  xi  by  a  —  xi. 

3.  5  _  2  V^l  by  3  +  2  V^l:  9.  a2  +  bH  by  a2 -  b2L 

4.  5  +  V^3  by  5  —  V3^.  10.  Vr  +  3iby  Vr  —  3i. 

5.  3-  V^by  3  +  2V:=~2.  11.  V^o  by  V^  by  V^5. 

6.  1  — V^7  by  2  +  3V^7.  12.  V— a  by  V— &  by  —  ci. 

III.   Division. 

Fractions  (that  is,  indicated  quotients)  may  be  simplified  by 
rationalizing  the  denominator  (Sec.  94,  p.  47). 

For  example : 

V3~7  =  V^T  V=T5  _  V7  •  V5  (~  I)2  =  V35  . 

V^~5      V^5  V^5  -  5  5 

2+  V3"3  _  (2  +  V^S) (3  +  V~^5)  _  6  +  3\/^3"+2\/^"5-vT5 

3+ V^5~(3_  V^5)(3  +  V^5)~  9  ~  (- 5) 

=  ^  (6  +  3  V^3 + 2  V^5"- VlT) . 

g  +  yi  _         (x  +  yQ2         _  a;2  +  2  syi  —  y2 
x  —  yi      (x-  yi)  (as  +  yi)  ~         x2  +  y2 

WRITTEN    EXERCISES 
Write  in  fractional  form  and  rationalize  the  denominators : 

7.  a-r-  (a  —  &i). 

8.  (a  +  bi)  -s-  (a  —  6t). 

9.  (3  +  6»)-i-(5+4t). 
10.    (V3-9i)-s-(V2-9i). 

11.  (x-V^Ty+ix+V^Ty 

13.    (V^-2H-V-5)-(V^r5"-V-2). 


1. 

V-6.-S-V2. 

2. 

1  -f-  (a  +  aa"). 

3. 

V-3-5- V-5. 

4. 

Vra-r  V —  a. 

5. 

l  +  (2-V-3). 

6. 

4V-l^--2V-4 

04  ELEMENTARY   ALGEBRA 

144.  Powers  of  the  Imaginary  Unit.     Beginning  with  r  =  —  1 
and  multiplying  successively  by  i  we  find : 

p  =  f2  .»•=•_£.  f  =  i6  •  i  =  —  1  ■  i  =  —  i. 

i*  =  i?.  r°=-l(-l)  =  +l.  i8=i*.i4=(+l)2=+l. 

i5  =  i4  •  j  =  i.  ?'9=  i'8  •  i  =  «'. 

145.  By  means  of  the  values  of  P,  i3,  i*,  any  power  of  i  can 
be  shown  to  be  either  ±  i  or  ±1. 

For  example  :  *63  =  j'60  ■  P  =  0'4)15  •  *'3  =  l15  •  *'3  =  «'3  =  -  *'• 

WRITTEN  EXERCISES 
Simplify  similarly : 

1.  /'•'.                   4.    i16.  7.    i:A.                 10.    ;'143. 

2.  i10.                   5.    r1.  8.    l56.                  11.    im. 

3.  P.                   6.    J-27.  9.    i198.                 12.    i3001. 

Perform  the  operations  indicated: 

13.  (1  +  ,)2  is.  (l-O3^.         it.  (1  +  0-* 

16.   (=1±**\\       is,  ^ 


14.  a-/)3.  v      2 

19.  (l  +  o  -(i-02-  20-  (i  +  o'-a-o2- 

IMAGINARIES   AS   ROOTS   OF   EQUATIONS 

146.    Complex  numbers  often  occur  as  roots   of   quadratic 
equations. 

EXAMPLE 

Solve :  x*  +  x  +  1  =  0.  (1) 

x*  +  x  =  -  1.  (2) 

Completing  the  square,  X2  +  X  +  ^  =  \  —  1.  (5) 

.■.3!  +  l=±V=l'.  (*) 

Test:    (-  \  ±  jV3  -i)2  +  (-  i  ± 'W5-0  +  1  =0. 


IMAGINARY   AND   COMPLEX  NUMBERS  95 


WRITTEN    EXERCISES 

Solve  and  test,  expressing  the  imaginary  roots  in  the  form 
a  +  hi : 

1.  a-2  +  5  =  0.  16.  12£2  +  24  =  0. 

2.  x*  +  2x  +  2  =  0.  17.  6w2+30  =  0. 

3.  x*  —  x  +  l=Q.  18.  8t2  +  t  +  6  =  0. 

4.  8*  +  x  +  5  =  0.  19.  7x2  +  x  +  5  =  0. 

5.  x2  +  2x  +  37  =  0.  20.  6or  +  3a;  +  l=0. 

6.  x2-Sx  +  25  =  0.  21.  4x2  +  4a;  +  3  =  0. 

7.  x2-  6  a;  +  10  =  0.  22.  12  a2  +  a;  +  1  =  0. 

8.  m2  +  4m  +  85  =  0.  23.  8 v2  +  3 v  +  6  =  0. 

9.  or2 +  10  a +  41=0.  24.  w2  +  5^o  +  6  =  0. 

10.  x2  +  30 .c  +  234  =  0.  25.  9z2  +  2z  +  5  =  0. 

11.  y2_4y  +  53  =  0.  26.  7  x2- 3  a +  4=0. 

12.  z2-6z  +  90  =  0.  27.  15z2  +  5z-l=0. 

13.  p2  +  20p  + 104  =  0.  28.  16ar-8a;  +  l  =  0. 

14.  2. r2 +  4^  +  3  =  0.  29.  10a? -2  a +  3  =  0. 

15.  3ar  +  2x  +  l  =  0.  30.  7f-t  + 1  =  0. 

147.  The  occurrence  of  imaginary  roots  in  solving  equations 
derived  from  problems  often  indicates  the  impossibility  of  the 
given  conditions. 

EXAMPLE 

A  rectangular  room  is  twice  as  long  as  it  is  wide;  if  its 
length  is  increased  by  20  ft.  and  its  width  diminished  by  2  ft., 
its  area  is  doubled.     Find  its  dimensions. 

Solution.     1.    Let  x  =  the  width  of  the  room,  and  2x  its  length. 

2.  Then  (2 x  +  20) (»- 2)  =2  •  2x  ■  x,  or*-  -  8x  +  20  =  0. 

3.  Solving  (2),  x  =  4  ±  2  i. 

The  fact  that  the  results  are  complex  numbers  shows  that  no  actual 
room,  can  satisfy  the  conditions  of  the  problem. 


96  ELEMENTARY   ALGEBRA 

WRITTEN    EXERCISES 

Solve  and  determine  whether  or  not  the  problems  are 
possible  : 

1.  In  remodeling  a  house  a  room  16  ft.  square  is  changed 
by  lengthening  one  dimension  a  certain  number  of  feet  and  by 
diminishing  the  other  by  twice  that  number.  The  area  of  the 
resultant  room  is  296  sq.  ft. ;  what  are  its  dimensions  ? 

2.  A  triangle  has  an  altitude  2  in.  greater  than  its  base, 
and  an  area  of  32  sq.  ft. ;  find  the  length  of  its  base. 

3.  A  train  moving  x  mi.  per  hour  travels  90  mi.  in  15  —  x 
hours.    What  is  its  rate  per  hour  ? 

SUMMARY 

I.  Definitions. 

1.  The  square  roots  of  negative  numbers  are  called  imaginary 
numbers.  Sec.  137. 

2.  In  distinction  from  imaginaries,  the  rational  and  irrational 
numbers  hitherto  studied  are  called  real  numbers.         Sec.  139. 

3.  A  complex  number  is  a  binomial,  one  of  whose  terms  is  a 
real  number  and  the  other  an  imaginary  number.  Sec.  141. 

II.  Processes. 

1.  After  introducing  the  symbol  i  for  the  imaginary  unit 
V— 1,  the  operations  with  imaginary  and  complex  numbers 
are  performed  like  the  operations  with  real  numbers. 

Sec.  142. 

2.  Any  power  of  i  can  be  expressed  by  ±  i  or  ±  1. 

Sec.  145. 

3.  The  solution  of  quadratic  equations  may  yield  complex 
numbers.  In  problems  this  often  indicates  the  impossibility 
of  the  given  conditions.  Sees.  146,  147. 


IMAGINARY   AND   COMPLEX   NUMBERS 


97 


SUPPLEMENTARY  WORK 

Graphical  Representation 

We  have  seen  that  positive  integers  and  fractions  can  be 
represented  by  lines. 

Thus,  the  line  AB  represents  3,  and     °. , 1 — *■  - — i — 


-+-» 


the  line  BC  represents  3£. 

Similarly,  we  have  seen  that  negative  integers  and  fractions, 
which  for  a  long  time  were  considered  to  be  meaningless,  can 
be  represented  by  lines. 

Thus,  the  line  BA  represents  —  3,  and 


<  i      ' ' 1    the  line  CB  represents  —  3|. 


Irrational  numbers  can  also  be  represented 
by  lines. 

Thus,  in  the  right-angled  triangle  abc,  the  line  db 
represents  the  y/2. 


I 

A 


x- 


t=& 


).!. 

B 


y 


;+1 


•~X 


Like  the  negative  number 
the  imaginary  number  re- 
mained uninterpreted  several 
centuries.  But  this  number 
also  can  be  represented  graph- 
ically. 


-i 


Thus,   if  a  unit  length  on  the 
y-axis  be  chosen  to  represent  V—  1 
J  or  i,  the  negative  unit  —  V—  1  or 

—  i  should  evidently  be  laid  off  in 
the  opposite  direction.     3V—  1  or 
3  i  would   then  be  represented  by 
OA  and  —  3  i  by  OB,  as  in  the  figure,  and  others  similarly. 

The  reason  for  placing  V  —  1  or  i  on  a  line  at  right  angles 
to  the  line  on  which  real  numbers  are  plotted  may  be  seen  in 


98 


ELEMENTARY   ALGEBRA 


the  fact  that  multiplying  1  by  V—  1  twice  changes  +  1  into 
—  1.  On  the  graph  +  1  can  be  changed  into  —  1  by  turning 
it  through  180°.  If  multiplying  1  by  V—  1  twice  turns  the 
line  1  through  180°,  multiplying  1  by  V—  1  once  should  turn 
+ 1  through  90°. 

For  example  : 

1.   Eepresent  graphically  V  — 4: 

V^4  =  vTi  =  2  i  i  this  is  represented  by  a  line  2  spaces  long  drawn 
upward  on  the  y-axis. 


2.    Eepresent  graphically  —  V— 3  : 

—  V—  3  =  —  V3  i  =  —  1.7  i  (approximately)  ;  this  is  represented  by 
a  line  1.7  i  spaces  long  drawn  downward  on  the  ?/-axis. 


WRITTEN    EXERCISES 


Represent  graphically : 


1.  3i. 

2.  -2L 

3.  V^9. 


4.    V-16. 


5.-5  i. 

6.  5  i. 

7.  V-^3 
8. 


9.    — 5V^4. 

10.  —31. 

11.  +2V^3. 


V-12. 


taken 


12.   5V-9. 

Complex  numbers 
may  be  represented 
graphically  by  a  modi- 
fication of  the  plan 
used  in  representing 
imaginary  numbers. 

EXAMPLES 

1.  Represent  graph- 
ically 3  +  i. 

To  do  this  3  is  laid  off 
on  the   axis  of   real   num- 
bers,  (vx'),  and   i  upward 
on  the  axis  of  imaginaries 
As  in  other  graphical  work  this  locates   the  point  Pi  which  is 
to   represent  the  complex   number,  3  +  i. 


IMAGINARY  AND  COMPLEX  NUMBERS  99 


The  number  Va'2  +  b-  is  called  the  modulus  of  the  complex  number 
o  +  M.  As  appears  from  the  figure,  OPi  =  V32  + 12,  and  hence  OPx 
represents  the  modulus  of  3  +  i. 

2.  Represent  graphically  3  —  i. 

The  point  P2   is  the  graph  of  the  complex  number  3  —  i,  and  OPg 

represents  its  modulus. 

3.  Represent  graphically  —  3  —  5  i. 

The  point  P3  is  the  graph  of  the  complex  number  —  3  —  5  f,  and  OP3 
represents  its  modulus. 

We  have  thus  interpreted  by  means  of  diagrams  positive  and 
negative  integers,  positive  and  negative  fractions,  positive  and 
negative  irrational  numbers,  and  positive  and  negative  com- 
plex numbers ;  in  fact,  all  of  the  numbers  used  in  elementary 
algebra. 


CHAPTER   VII 

QUADRATIC   EQUATIONS 

GENERAL  FORM 

148.  The  general  form  for  a  quadratic  polynomial  with  one 
unknown  quantity  is  ax2  +  bx  +  c,  where  a,  b,  and  c  denote  any 
algebraic  expressions  not  involving  x,  and  where  a  is  not  zero. 
If  a  is  zero  the  polynomial  is  linear. 

For  example:  1.   5x2  — 7x  +  8. 

Here  a  =  5,  b  =  —  7,  c  =  8. 

»        7m-x*  +  3x         5m 


2n  +  l  2w-l 

HereQ  =  ,7w    ,   6  =  3,   c  =  ~5m 


2b  +  1  2  » - 1 

WRITTEN    EXERCISES 
Put  the  following  expressions  into  the  form  ax2  +  &x  +  c: 

1.  3x  +  5x(x-2)  +  4:(x2-5).  4.    (a;  +  9)  -  q(s*  - 11). 

2.  7(4  x—  1) +  0  +  3)(x  —  2).  5.    (a&  +  &)(ca>  +  d). 

3.  a(6x  +  c)(2dx  +  3e).  6.    (x2  —  a)  +  (a*  -  b). 

7.    (x  +  g)(x+p)-(x-g)(2x-^). 


_     fx,l\*     2/9  x     16V 
8-    ^  +  3j~3(,T~"3j 


9.   x2  +  a&  —  ax  —  b(a  +  x  +  x2). 

10.  x(x-2)(x-4)-x-°(x-5). 

11.  (2x  +  l)2-(3x  +  l)2+(4x  +  l)2. 

12.  (x  -  1)0  -  2)0  -  3)  -0  + 1)0  +  2)0  +  3). 

100 


QUADRATIC   EQUATIONS  101 

149.  Similarly,  every  quadratic  equation  can  be  put  into  the 
form  ax2  +  bx  +  c  =  0  by  transposing  all  terms  to  the  left  mem- 
ber and  then  putting  the  polynomial  which  constitutes  the  left 
member  into  the  form  ax2  +  bx  +  c. 

EXAMPLES 

1.  (3x  +  5)(2x-7)  =  3x2-4, 
then  6x2  -  11  x  -  35  =  3 a2  -  4, 

or,  3x2-llx-31  =  0. 
Here  a  =  3,  b  =  -  11,  c  =  -  31. 

2.  (??ix  +  3  a)2  =  mx2  -  5  (amx  -  2), 

then  m2x2  +  6  ami  +  9o2  =  mx2  —  5  amx  +  10, 

or,    (m2  -  m)x2  +  11  amx  +  9  a2  —  10  =  0. 
Here  a  =  m2  —  m,  &  =  11  am,  c  =  9  a2  —  10. 

WRITTEN   EXERCISES 
Put  the  following  equations  into  the  form  ax2  +  bx  +  c  =  0 : 

1.  (a-l)8  =  (aH-l)3.  3.    (7x-±)2  =  3z  +  2. 

2.  a;2  +  ex  =  /a  +  g(x  +  e).  4.    (5  a  ~  ^  =  18  a  -  2  a. 

X         ,       \2        /      X  ^2 

+a    = —a 


a+1       J      \a— 1 
_     a-8x     a  —  4  «     a  —  5  a; 

b.     — —  = •   • 

a  +  6x     a  —  3x     a  +  5x 
7.    (x  +  l)(x2-l)  =  (x2  + 1) (a;  +  2). 

8    (  -       3a\2_f  x       3a\fx     4' 
V2«       4/\2&     Ty\5     9 

_     2a  +  5&4-3z     «  +  &      3a-6  +  2a; 
y.    = •  • 

3a  —  5  6  +  3  #     a— &     2a  —  b  +  3x 

10.  (2  a  -  4  +  a)2  +  4(a  4-  4  +  a;)2  =  (3  a  +  6  +  2  x)\ 

11.  (2a;-|-4&-3)2+(2®  +  2&  +  ll)2 

=  (a;  +  3  b  -  8)2  +  (3  x  +  3  b  +  8)2. 


102  ELEMENTARY   ALGEBRA 


METHODS  OF  SOLUTION 

150.  General  Solution.  By  solving  the  general  quadratic 
equation  ax2  +  bx  +  c  =  0,  general  formulas  for  the  roots  are 
obtained. 

Solve:  ax2  +  bx  +  c  =  0.       (1) 

Dividing  by  a,  which  is  not  0,  X2  +  —  +  -  =  0.  (~) 

a      a 

Adding  -^—  to  complete  the 

4fl2  hr         7)2  /,2         ,. 

square  and  subtracting  the  x2+—  +  — - V  -  =  0.  (5) 

same,  a        4  ,/2       4  f<2       a 


Writing  the  second  term  as 
the    square   of   its   square  ,  ?    ,2       /  y/frl  _  4  ac\  2 

Factoring  (5), 


\       2a  2a        j\        2a  2a       )  K  ' 

2a     V        2a        /  2a     \        2a       J  K  l 

Denoting  these  roots  by  rx  and  r2 : 


-f>  +  V&2-4ac 
ri  = 2a 


—  b—  V&2  —  4  ac 

**2  = ~ 

2a 
By  substituting  in  these  formulas  the  values,  including  the 
signs,  that  a,  b,  c  have  in  any  particular  equation,  the  roots  of 
that  equation  are  obtained.     This  is  called  solution  by  formula. 

EXAMPLE 

Solve :  3  x2  -  9  x  +  5  =  0.  (1) 

Here,  a  =  3,  b  =  -  9,  c  =  5,  (2) 


_  9  ±  V81  -  m  _  0  ±  V2T  ... 

6  (j 


QUADRATIC   EQUATIONS  103 

WRITTEN    EXERCISES 

Solve  by  formula : 

1.  x2- x  -1  =  0.  9.  x2-x  +  6  =  0. 

2.  ar  +  3.r  +  l  =  0.  10.  2a2  —  as  +  2  =  0. 

3.  ar  +  2.r-l  =  0.  11.  3.C2- 2  a;  +  1  =  0. 

4.  x2- 4  a  4- 4  =  0.  12.  7^  +  6a;-4  =  0. 

5.  or9-  5 x  +  6  =  0.  13.  4  a,-2 -12  x  +  9  =  0. 

6.  ^-3.r  +  2  =  0.  14.  3x2  +  5x-2  =  0. 

7.  a-2-13a-  +  9  =  0.  15.  5  a2- 4x  + 6  =  0. 

8.  2x2-7x-3  =  0.  16.  7x2  +  5x-8  =  0. 

151.  Literal  Quadratic  Equations.  When  any  of  the  coeffi- 
cients of  a  quadratic  equation  involve  letters,  the  equation  is 
called  a  literal  quadratic  equation. 

Such  equations  are  solved  in  the  usual  way. 

EXAMPLES 

1.  Solve:  x2 4- 6  mx  +  8  =  0.  (1) 

x2  +  6mx=-8  (£) 

Completing  the  square,    X2  +  6  mx  +  9  m2  =  9  m2  —  8.  (3) 

.-.  (x  +  3  m)2  =  9  roa  -8.  (4) 

.-.  x  +  3ct  =  ±  V9m2-8.  (5) 

.-.  x  =  -  3  m±  V9  m2  —  8.  (6) 

2.  Solve :  £2  +  gt  +  /i  =  0.  (i) 

Here  0  =  1,      6  =  g,      C  =  h.  (#) 


Hence,  bv  Sec.  150,  t-~9  ±  ^92  ~  +  h . 

2 


(3) 


3.    Solve:                          gt2  +  2vt  =  2s.  (1) 

gt2  +  2vt-2s  =  0.  (2) 

Here                     a  =  g,     b  =  2  w,     c  =  —  2  s.  (3) 

Hence,  by  Sec.  150,  t  =  -  —  ±  —  V(2  I')2  -  4  gr  f-2s) .        (4) 

2<7      2g 

=  -  -  ±  -  Vv2  +  2gs.  (5) 

a    g 

=  -(-v±Vv2  +  2gs).  (£) 


104 


ELEMENTARY   ALGEBRA 


WRITTEN    EXERCISES 


Solve : 

1.  f-+at  =  7c. 

2.  u2  4-  ku  + 1  =  0. 

3.  v2  +  mo  =  1. 

4.  ax2  +  bx  +  c  =  0. 

5.  a:24-ax  +  o  =  0. 

6.  m2^2  +  2  ma;  =  —  1. 

7.  x~  +  2px-l  =  0. 

8.  4 a^-4ax  +16=0. 

9.  aV  4- 2  ax  4- 5  =  0. 
10.  m¥  +  4mx-6=0. 


11.  x2-4ax  =  9. 

12.  £2-8*4-24d  =  9d'2. 

13.  5aarJ4-3fre  +  2&3  =  0. 

14.  ay2-(a-b)y-b  =  0. 

15.  62a^  -  2  to  =  ac  - 1. 

16.  w2  4- 4  aw  4- a2  =  0. 

17.  .T2-3«x-(-10a2  =  0. 

18.  v2  —  4;  amv  =  (a2  —  m2)2. 

19.  2arJ-3a  =  a(3-4a;). 

20.  w2-a2=2b(a-  w). 


152.    Collected   Methods.    We   have  used   three    methods  of 
solving  quadratic  equations : 


1.    Factoring. 

Equation 

^-3x4-2  =  0. 

x2  —  (a  +  b)x  +  ab  =  0. 


Factors 

(x~2)(x-l). 
(x  —  a){x  —  b). 


Roots 
X  =  2,  X  =  1. 

x  =  a,  x  =  b. 


2.    Completing  the  square. 

Equation  Solution 

cc2  +  x  +  2  =  0.         See  First  Course, 

Sec.  285,  p.  225. 

ax2  +  bx  +  c  =  0.  See  Sec.  150. 


Roots 


£C  = 


X: 


-l±V-7 


&±V&2-4 


ac 


3.   Formula. 

Equation 

3ar  +  2ic-7  =  0. 


Solution 

See  Sec.  150. 


ax2  +  bx  +  c  =  0.        See  Sec.  150. 


iC  = 


#  = 


2a 

Roots 

-1±V22    . 

—  ■  ■  • 

3 

—  b±  -y/b2—  4  ac 
2a~~ 


QUADRATIC   EQUATIONS  105 


WRITTEN    EXERCISES 


6. 

a^_a;_30  =  0. 

7. 

x2  +  x-12  =  0. 

8. 

x2  -  3  x  +  2  =  0. 

9. 

x2  + 11  x  +  30  =  0. 

0. 

x2  -  7  x  +  12  =  0. 

Solve  by  factoring : 

1.  x2-x-6  =  0. 

2.  x2-x-2  =  0. 

3.  £2  +  £-2  =  0. 

4.  k2  +  jc-6  =  0. 

5.  ar9  +  3£  +  2  =  0. 

Solve  by  completing  the  square : 

11.  x*  +  x  +  l  =  0.  15.  x2  -  5x  +  10  =  0. 

12.  ar  +  3£+l=0.  16.  x*- 16 x  +  60  =  0. 

13.  o2-^  -|-1  =  0.  17.  ar'  +  f  ic  -h  i  =  0. 

14.  ^-.9x-+.5  =  0.  18.  a? +  1.5  a; -3.5  =  0. 

Solve  by  formula : 

19.  3£2  +  £  +  5  =  0.  24.  x2  +  l  =  0. 

20.  2^ -5 £-3  =  0.  25.  x2  + 15  x  +  56  =0. 

21.  4  a;2  +  3  x  -1  =  0.  26.  x2  +  8  x  +  33  =  0. 

22.  5ar  +  2z  +  6  =  0.  27.  a2- 10 x  +  34  =  0. 

23.  £2  +  £  +  l  =  0.  28.  2 a^  +  3 £-27  =  0. 

Solve  and  test,  using  whichever  of  the  methods  in  Sec.  152 
seems  most  convenient :  • 


29. 

9  y2  -  4  =  0. 

30. 

Qx2-  13  £  +  6  =  0 

31. 

5  as2  —  4  £  +  4  =  0. 

32. 

f  + 11 1  +  30  =  0. 

33. 

6«*-5s-6  =  0. 

34. 

6r2-2r-4  =  0. 

35. 

ws  +  4  w>  -  3  =  0. 

36. 

2£-l      2.7- +  1_ 

2£+l        2£-l 

8 

37. 

£2-2x  +  3  =  0. 

38. 

x2-  0.3  a»  +  0.9  =  0. 

39. 

aJS_  l.i  x  +  1.2  =  0. 

40. 

11  0^  +  1  =  4(2- £)2. 

41. 

x2+(a  +  b)x  +  ab  =  0. 

42. 

£2-(6  +  c)£  +  6c  =  0. 

43. 

2a£  +(a-2)£-l  =  0 

44. 

9                           2 

6  +  £        6  —  £ 


106  ELEMENTARY   ALGEBRA 

45.  -2+-^ +  _*<). 

x  —  4      a;  —  6      x  —  2 

46.  i(x-l)(x-2)  =  (x-2%)(x-l§). 

47.  The  product  of  two  consecutive  positive  integers  is  306. 
Find  the  integers. 

Solution. 

1.  Let  x  be  the  smaller  integer. 

2.  Then  x  +  1  is  the  larger. 

3.  .-.  x(x  +  1)  is  their  product. 

4.  .-.  x(x  +  1)  =  306,  by  the  given  conditions. 

5.  .-.  x2  +  x  -  306  =  0,  from  (4). 

6.  ...  x  =  -l±Vl  +  1^4  =  -1±35  =  17?  or  _  18>  golving  (5) 

Since  the  integers  are  to  be  positive,  the  value  —  18  is  not  admissible. 

*  =  17,  /.  x  +  1  =  18,  and  the  integers  are  17  and  18. 

Test.     17  •  18  =  306. 

48.  There  is  also  a  pair  of  consecutive  negative  integers 
whose  product  is  306.     What  are  they  ? 

49.  If  the  square  of  a  certain  number  is  diminished  by  the 
number,  the  result  is  72.     Fiud  the  number. 

50.  A  certain  number  plus  its  reciprocal  is  —  2.  What  is  the 
number  ? 

51.  A  certain  positive  number  minus  its  reciprocal  is  f-. 
What  is  the  number  ?  What  negative  number  has  the  same 
property  ? 

•  52.  One  perpendicular  side  of  a  certain  right  triangle  is  31 
units  longer  than  the  other;  the  square  of  their  sum  exceeds  the 
square  of  the  hypotenuse  by  720.    Find  the  sides. 

53.    The    perimeter    of   the    rectangle 
x  shown  in  the  figure  is  62  in.     Find  the 
sides. 

54.  In  a  right  triangle  of  area  60  sq.  ft.  the  difference 
between  the  perpendicular  sides  is  7.     Find  the  three  sides. 

55.  The  sum  of  the  hypotenuse  and  one  side  of  a  right 
triangle  is  162,  and  that  of  the  hypotenuse  and  the  other  side 
is  121.    What  are  the  sides  '.' 


QUADRATIC    EQUATIONS  107 

RELATIONS   BETWEEN   ROOTS   AND   COEFFICIENTS 

153.  Relation  of  Roots  to  Coefficients.  By  adding  and  multi- 
plying the  values  found  for  the  roots  (Sec.  150),  we  obtain 
respectively, 

fi  +  r,  — jL 

c 

Applying  this  result  to  the   equation   x?+px-\-  q  =  0,   we 

have: 

n  +  r2  =  -p, 

»i  r2  =  q- 

In  words : 

In  the  equation  x2+px  +  q  =  0,  the  coefficient  of  x  ivith  its  sign 
changed  is  the  sum  of  the  roots,  and  the  absolute  term  is  their 
product. 

Every  quadratic  equation  can  be  put  into  the  form  x'1  +  px  +  q  =  0  by 
dividing  both  members  by  the  coefficient  of  x'1. 

154.  By  means  of  Sec.  153  a  quadratic  equation  may  be 
written  whose  roots  are  any  two  given  numbers. 

EXAMPLES 

1.  Write  an  equation  whose  roots  are  2,  —  3. 

_^  =  n  +  r2  =  2+(—  3)  =  —  1.     .-.p  =  l 

q  =  nr2  =  2(—  3)  =  —  6. 
.-.  x2  +  x  —  6  =  0  is  the  equation  sought. 

2.  Write  an  equation  whose  roots  are  \  +  V  — 3,  ^  —  V  —  3. 

-p  =  n  +  r2  =  a  +  V^=3)  +  (£-  V^3)=  1.     .-.p  =  -  1 
q  =  IV,  =  (J  +  V^3) (i  -  V^3)  =  i -(- 3)  =  V- 
.• .  a:'2  —  x  +  A£  =  0  is  the  equation  sought. 

WRITTEN    EXERCISES 
Write  the  equations  whose  roots  are : 

1.  4,  5.  3.   24,  30.  ,  5.    a,  -b. 

2.  f,  f .  4.   8|,  10.  6.   8,  -  40. 


108  ELEMENTARY  ALGEBRA 

7.  7,  -If  10.    —5.  —20.  13.   a  —  bi,  a  +  bi. 

8.  _4,  +  4.         11.    -f±iV5.        14.    1  +  2/,  1-2 1. 

9.  |±V=6.       12.   |±|V^47.      15.   i-V2,  i+V2. 

155.  Testing  Results.  The  ultimate  test  of  the  correctness 
of  a  solution  is  that  of  substitution ;  but  this  is  not  always 
convenient,  especially  when  the  roots  are  irrational.  In  such 
cases,  the  relations  between  the  roots  and  coefficients  are  of  use. 

For  example  :    Solving  2  a;'2  —  5  x  +  6  =  0, 

or  x2  —  |  x  +  3  =  0,  the  roots  are 


n  =  |  +  \  V-  23  and  r2  =  f  -  \V^  23. 

Adding,  —  (r\  +  r2)  =  —  -¥°  =  —  f ,  the  coefficient  of  x. 


Multiplying,  n  r2  =  (f)2  -  Q V-  23)*  =  §f  +  ff  =  3, 
the  absolute  term. 

Therefore,  the  roots  are  correct.     (Sec.  153.) 

156.    In  what  follows,  the  coefficients  a,  b,  c,  are  restricted 
to  rational  numbers. 


157.    Character  of  the  Roots.     By  examining  the  formula  for 

the  roots, =-^- — — — — ,  it  appears  that  the  character  of  the 

2  a 

roots  as  real  or  imaginary,  rational  or  irrational,  equal  or  un- 
equal, depends  upon  the  value  of  the  expression  b2  —  4  ac. 

1.  If  b2  —  4  ac  is  positive,  the  roots  are  real. 

Thus,  in  x2  +  4  x  -  3  =  0,  b2  —  4  ac  =  16  +  12,  or  28,  .-.  the  roots  are 
real  and  unequal. 

2.  If  62  —  4  ac  is  a  perfect  square,  the  indicated  square  root  can  be 
extracted,  and  the  roots  are  rational. 

Thus,  in  x2  -  4  x  +  3  =  0,   b2  -  4  ac  =  16  —  12,  or  4,   .-.  the  roots  are 
rational  and  unequal. 

3.  If  b2  —  4  ac  is  not  a  perfect  square,  the  indicated  root  cannot  be 
extracted  and  the  roots  are  irrational. 

Thus,    in   x2  +  5x  +  1  =  0,    b2  -  4  ac  =  25  -  4  =  21,    .-.  the   roots  are 
irrational. 


QUADRATIC    EQUATIONS  109 

4.  If  62  —  4  fflc  =  0,  the  radical  is  zero,  and  the  two  roots  are  equal. 

Thus,  x2  -  lOx  +  25  =  0,  b2  -  4  ac  =  100  -  4  •  25  =  0,  .-.  the  roots  are 
equal. 

5.  If  b2  —  4  ac  is  negative,  the  roots  are  imaginary. 

Thus,  in  2  x2  —  x  +  1  =  0,  b2  —  4  ac  =  1  —  8,  or  —  7,  .*.  the  roots  are 
complex  numbers. 

Consequently,  it  is  merely  necessary  to  calculate  b2  —  4  ac  to  know  in 
advance  the  nature  of  the  roots  of  a  quadratic  equation. 

158.  Discriminant.  Because  its  value  determines  the  char- 
acter of  the  roots,  the  expression  b2  —  4  ac  is  called  the  dis- 
criminant of  the  quadratic  equation. 

ORAL    EXERCISES 
Without  solving  the  equations,  find  the  nature  of  the  roots  of : 

l.   a?  +  x-20  =  0.  9>   l_+lsaXm 

4:X 


2.  x*  +  x-3  =  0. 

3.  2x2-x  +  2  =  0. 

4.  3x2-x-\-3  =  0. 


10.  7ar>  +  3a;-4  =  0. 

11.  _4cc  +  8x2  +  l  =  0. 

12.  5  +  4x-2-3a:  =  0. 

5.  2s»  +  2s~4  =  0.  13    7aj+6  +  aj2  =  0. 

6.  5aj2-3a;  +  6=0.  14     _6o;  +  9^  +  3  =  0. 

7.  3x-2-4z  +  5  =  0.  15.   £c2-6a;  +  4  =  0. 

8.  6 z2  +  a-  1  =  0.  16.   5^-1  +  ^  =  0. 

159.  The  relation  x2  -\-px  +  q  =  x2  —  (i\  +  r2)  x  +  rfa  may  be 
written : 

(1)  x2  +  px  +  q  =  (x  —  n)  (fl5  —  r2). 

And  since  x*+px+  q  =  °^  +  bx  +  c  [n  ^oh.  p  =  &  ?  q  __£. 

a  a  a 

we  have 

(2)  ax2  +  to  +  c  =  a  (a  —  ?-,)  (a  —  r2). 

160.  The  solution  of  a  quadratic  equation,  therefore,  enables 
us  to  factor  every  polynomial  of  either  form  (1)  or  (2). 


HO  ELEMENTARY   ALGEBRA 

Since  i\  and  r2  involve  radicals  : 

1.  Tlte  factors  mil  generally  be  irrational. 

2.  Tlie  factors  will  be  rational  when  ra  and  r2  are  so  ;  that  is, 
when  b2  —  4  ac  is  a  perfect  square. 

3.  Tlie  tiro  factors  involving  x  icill  be  the  same  when  the  roots 
are  equal ;  that  is,  when  b2  —  4  ac  =  0. 

In  the  last  case  the  expressions  are  squares  and 

(1)  becomes  (x  —  i\)2,  and 

(2)  becomes  [-\/a(x  —  r^]2. 

EXAMPLES 

Trinomial  W  -  4  ac  Nature  of  Factors 

1.  3x2  — 7x  +  2  49  —  4.3-2=25  rational  of  1st  degree. 

2.  3  x2  -  7  x  +  3  49  -  4  •  3  •  3  =  13  irrational. 

3.  2x2-8x  +  8  64-4-2-8  =  0  equal. 

ORAL  EXERCISES 

By  means  of  the  above  test,  select  the  squares ;  also  the 
trinomials  with  rational  factors  of  the  1st  degree : 

1.  8  a2 -8  a; +  2.         5.   a2 +  3  a; -2.  9.    6  a-2 +  5  a:  -4. 

2.  ^  +  4«  +  12.         6.   a?x2  +  2ax+l.     10.    6  x2  —  5x  +  9. 

3.  3^  +  3flJ  +  l.         7-    4ar°  +  4x-  +  l.       n.    4ar-4a;-3. 

4.  3z2  +  2z  +  12.        8.    a--2 -8  a; +  15.        12.   8X2  -9  x  +  3. 

161 .    The  actual  factors  of  any  quadratic  trinomial  of  the  form 
ax2  +  bx  +  c  can  be  found  by  solving  the  quadratic  equation : 

ax2  +  bx  +  c  =  0, 
and  substituting  the  roots  in  the  relation : 

ax2  +  bx  +  c  =  a  (x  —  r2)  (as  —  r2). 

EXAMPLE 
Factor:  6  a;2 +  5  a; -4.  CO 

Solving  6*2  +  5* -4  =  0,  X  =  —  f,  (~) 

Substituting -I  for  r,     a(x  _  ri)(a;  -  r2)  =  6(x  +  f)(x  -  £)•      (5) 
5  for  r2  and  ti  for  a. 

Therefore,  6  X2  4-  5  X  -  4  =  6  (x  +  f)  (x  -  J) .  (4) 


QUADRATIC   EQUATIONS 


111 


Factor : 

1.  3  a:2 -2  a; -5. 

2.  9  a2 -3  a; -6. 

3.  Qf+y-1. 


WRITTEN    EXERCISES 

5.  10w2-12w+2.      9.  6  a2  -7  x  +  3. 

6.  9u2-17u-2.      10.  5  x2- 40  a; +  6. 

7.  6  x2  +  25  a;  +  14.    11.  a2"* -2  a" -3. 


4.   15 y*-4Ly -35.    8.   2z2  +  5z  +  2.        12.   c4- 13  c2 +  36. 


GRAPHICAL  WORK 

162.  Preparatory. 

1.  By  counting 
spaces  read  the  length 
of  EF  in  the  figure. 

2.  Is  it  the  square  of 
the  length  of  OE  ? 

3.  Answer  similar 
questions  for  GH  and 
OH. 

Every  point  of  the 
curve  is  so  located  that 
the  length  of  its  ordinate 
is  the  square  of  its  ab- 
scissa. 

163.  Quadratic  ex- 
pressions may  be  repre- 
sented graphically. 

For  example  : 

The  curve  in  the  figure 
is  the  graph  of  y  =  x?. 
That  is,  the  length  of  CD 
is  the  square  of  that  of 
OC  ;  the  length  of  AB  is  the  square  of  that  of  OA  ;  etc. 


i 

1 1 ] 

/ 

! 

■••; 

I L..16 

IS   J 

1      -; 

1 

/  : 

1 :1 j 

: : 

\ 

/                     ■■ 

I ;■■  I 

; 

-••! f 

9 

[F 

••••j i 

\ 

-] I 

4 

/b 

!        ; 

*  -                 :                 : 

1 

■4 ! ■•■- 

\        1 

h 

x1'      : 

y 

c 

A 

E 

!h    L 

0 

1 

2 

i 

14 

y 


NlWIHEK 

SQUARE 

-  5 

25 

-4 

16 

-3 

9 

-2 

4 

-  1 

1 

-0 

0 

1 

1 

2 

4 

3 

9 

4 

16 

5 

25 

112  ELEMENTARY  ALGEBRA 

WRITTEN    EXERCISES 

1.  Construct  on  a  large  sheet  of  squared  paper  the  points 
corresponding  to  this  table  of  squares. 

2.  Then  sketch  a  smooth  curve  through  the 
points  beginning  with  —  5,  25. 

The  work  should  be  carefully  done,  and  the  result  pre- 
served for  later  use.  As  there  are  no  negative  values 
of  x2,  the  x-axis  should  be  taken  near  the  lower  edge  of 
the  paper.  The  unit  should  be  chosen  quite  large  ;  for 
example,  10  spaces.  Then  the  table  might  include  squares 
of  numbers  increasing  by  tenths:  1,  1.1,  1.3,  etc.  The 
curve  will  be  a  graphical  table  of  squares  and  square 
roots. 

3.  Read  to  one  decimal  place  from  the  graph 

V2;    V3;  V5;   V6;  V7;   V8 

Every  ^-distance  is  the  square  of  the  corresponding  x-distance  ;  and 
every  x-distance  is  the  square  root  of  the  corresponding  ?/-distance.  We 
see  that  for  every  y-distance  there  are  two  corresponding  x-distances, 
one  plus  and  the  other  minus,  corresponding  to  the  two  square  roots. 
Thus  the  points  of  the  curve  for  which  y  =  4  are  those  whose  values 
of  x  are  2  and  —  2  respectively,  i.e.  Vi  =  ±  2. 

164.  Graphical  Solution  of  Quadratic  Equations.  Any  value  of 
x  which  satisfies  the  system 

\y  =  -px-q 

makes  x2  equal  to  —  px  —  q,  or  x2  -\-px  +  q  =  0. 

The  values  of  x  satisfying  the  system  may  be  read  from  the 
graph  of  y  =  x2. 

EXAMPLE 

Solve  graphically  x2  —  x  —  6  =  0. 

1.  Construct  the  graph  of  y  =  x2.    (As  in  Sec.  163.) 

2.  Construct  the  graph  of  y  =  x  +  6. 

3.  They  intersect  at  points  for  which  x  =  —  2  and  +  3. 
.  ••  the  roots  of  x2  —  x  —  6  =  0  are  —  2,  3. 


QUADRATIC   EQUATIONS 


113 


Notes.     1.    Step  2  may  be  done  by  simply  noting  two  points  of  the 
graph  of  y  —  x  +  6  and  laying  a  ruler  connecting  them.     The  roots  can 


be  read  while  the  ruler  is  in  position,  and  thus  the  same  graph  for  y  =  x2 
can  be  used  for  several  solutions. 

2.    The  equation  must  first  be  put  in  the  form  x2  +  px  +  q  =  0,  if  not 
so  given. 

WRITTEN    EXERCISES 

Solve  graphically  : 

1.  x2-5x  +  §  =  0. 

2.  x2  +  3  x  +  2  =  0. 

3.  .r-2x-3=0. 

4.  ^+2^-3=0. 

5.  ar-3x-40  =  0. 

6.  z2  +  4;c+4  =  0. 


7.  2^-^-1  =  0. 

8.  3a2-2a-l=0. 

9.  x2  +  x  +  ±  =  0. 

10.  x2  +  x—2=0. 

11.  4f  +  4.r  +  l=0. 

12.  a2- 9  =  0. 


13.    The  path  of  a  projectile  fired  horizontally  from  an  eleva- 
tion, as  at  0  in  the  figure  on  p.  114,  with  a  given  velocity  may 


114 


ELEMENTARY   ALGP^BRA 


be  represented  by  the  graph  of  the  equation  y  =  f—  ,  where 


2v2' 


g  =  o2  and  v  is  the  initial  velocity  of  the  projectile  in  feet 
per  second.  Let  v  =  16  ft.  per  second  and  compute  the  num- 
bers to  complete  the  table  of  values  of  x  and  y. 


Table 


1        7      ;3      ;4      :5      16      [7      \8      \9      MO 


X 

y 

0 

0 

1 

( ) 

4 

( ) 

8 

( ) 

9 

( ) 

16 

( ) 

24 

( ) 

32 

( ) 

48 

( ) 

Read  from  the  graph  of  this  table  the  horizontal  distance 
traveled  by  the  projectile  when  it  is  4  ft.  below  the  starting 
point. 

14.  Construct  similarly  the  path  of  a  projectile  whose  initial 
velocity  is  32  ft.  per  second. 

15.  A  cannon  of  a  fort  on  a  hill  is  300  ft.  above  the  plane 
of  its  base.  The  cannon  can  be  charged  so  as  to  give  the 
projectile  an  initial  velocity  of  100  ft.  per  second.  What  range 
does  the  cannon  cover  ? 

16.  The  enemy  is  observed  at  a  point  known  to  be  1\  mi. 
from  the  foot  of  the  vertical  line,  in  which  the  cannon  stands. 
With  what  initial  velocity  must  the  ball  be  discharged  to  strike 
the  enemy  ? 


QUADRATIC   EQUATIONS  115 

CERTAIN   HIGHER   EQUATIONS   SOLVED  BY  THE  AID  OF 
QUADRATIC   EQUATIONS 

165.  We  have  found  the  general  solution  of  linear  and 
quadratic  equations  with  one  unknown.  Equations  of  the 
third  and  the  fourth  degree  can  also  be  solved  generally  by 
algebra,  and  certain  types  of  equations  of  still  higher  degree 
as  well ;  but  these  solutions  do  not  belong  to  an  elementary 
course.  We  shall  take  up  only  certain  equations  of  higher 
degree  whose  solution  is  readily  reduced  to  that  of  quadratic 
equations. 


CO 

(*) 

(3) 
(4) 
(5) 

(«) 

The  four  values,  of  x  are  the  four  roots  of  the  given  equation  of  the 
fourth  degree.     Test  them  all  by  substitution. 

2.  Solve:  x6- 3  ^-4  =  0.  (1) 

Let  y=  a-s,  then         y2  —  3  y  —  4  =  0.  (2~) 

Solving  (2),  y  —  4, 

and  y  ="  -  1.  '   (5) 

.'.  by  the  substitution  in  (2),        Xs  =  4,   or  X3  —  4  =  0, 

and  X'  -  —  1,  or  X3  +  1  =  0.  {4) 

Factoring  (A),  Xs  -  4  =  (x  -  ^I)(X2  +  Vi  ■  X  +  V¥)  =0.  (5) 

and  X3  +  1  =  (X  +  1)(X2  — X+I)=0.  (6) 

Solving  (j),  x=</i,  or  #4(_.|+jV^3),  or  ^4(-i-£V=3).     (7) 

Solving (6),  x=-l,  or  ±— |\/^3,  or  i  +  i\/^3).  (5) 

3.  Solve:       (ar8  -  3  a; +  l)(arJ- 3 x  +  2)  =  12-  CO 

This  may  be  (x2  _  3  x  +  nr(x2  _  3  X  +  1)  +  1]  =  12.  (2) 

written 
If  y  is  put  for 

sb*  — S.a>+1,  the 

equation  becomes  2/(2/  +  1)  =  12,  (o) 

or  J/2  +  y  -  12  =  0.  (4) 


EXAMPLES 

Solve : 

.r4- 

9  x-2  +  8  =  0. 

Let  y  =  .v2 ;  then  the  given 

equation  becomes, 
Solving  for  y, 

,,2 

-92/+8  =  0. 

y  =  8  or  1. 

Therefore 

x2  =  8. 

Or, 

x2  =  l. 

Solving  (A),  (5), 

x  =  ±  V8,  ±  1 

116  ELEMENTARY   ALGEBRA 


Solving, 

y  =  3  or  ■ 

-4. 

(5) 

Then  from  (3), 

x-  -  3  *  +  1  =  3, 

(6) 

and 

x2-3x  +  1  =-4. 

(7) 

Solving  (6), 

x  =  S±VTf- 
2 

(*) 

Solving  (7), 

x      3±V-11 

(9) 

These  are  the  four  roots  of  the  given  equation  of  the  fourth  degree. 

WRITTEN    EXERCISES 
Solve  as  above  : 

1.  x«-7a?+6  =  0.  9.    a-3  =  l-a:6. 

2.  rf-8-M-2=0.  1Q>   ^  +  5=     5 


3.  a;10 -5. ^  +  6  =  0.  ^  +  3 

4.  x4  +  13x2  +  36  =  0.  11.  x2n-Axn-5  =  0.    , 

5.  *»-3a?+l  =  0.  12.  2a:6  +  5ar!  +  2  =  0. 

6.  12  -  x*  =  11  x-2.  13.  se*  +  aa?-8a2  =  0. 

7.  af"  -  6icn  +  c  =  0.  14.  (a;2  +  4)2-4(a:2+4)+4=0. 

8-  -it—. r  +  -A^  =  V^ -  15.  a2+3a;=l 


z2  +  l      a2 +  2      a2 +  3  -r".  -        g^  +  Sx  +  1 

16.  (a?-3x  +  l)(a?  —  3x  +  2)  =  12. 

17.  (a;2-l)2  +  2(x-2-l)  +  l  =  0. 

18.  (x2  +  5x  —  l)(ar  +  5a  +  l)=  —  1. 

166.  Binomial  Equations.  Equations  of  the  form  xn±a  =  0 
are  called  binomial  equations.  The  simpler  cases  admit  of  being 
solved  by  elementary  processes. 

EXAMPLES 

1.    Solve  :  x>  -  1  =  0,  or  x3  =  1.  (7) 

Factoring  ai»- 1,  (a;  —  1)  (x2  +  X  +  1)  =  0.  (0) 

Finding  equations  .  . 

equivalent  to  (2),      X  =  1,  X2  +  £  +  1  =  0.  (3) 

_  1  _  ■«/_  3 
Solving  (3),  X  =  1,  X  = (4) 

Thus  we  have  found  the  three  numbers  such  that  the  cube  of  each  is  1, 
or  the  three  cube  roots  of  unity. 

Verify  this  statement  by  cubing  each  number  in  step  (4). 


QUADRATIC   EQUATIONS  117 

Note  that       (  -  I  +  i  V^3)2  =  ~  1  ~ V~  3  =  ~1~'V§, 

2t  2t 


1      *i    4.              f      i       i    / — 5\o       —  1  +  V—  3      —  1  +  i  V3 
also  that  (-i~iv  — 3)2= ±- = ^ 

Hence,  if  w  stands  for  one  of  the  complex  cube  roots  of  unity,  w2  is  the 
other. 

Every  number  has  three  cube  roots  ;  for  example,  the  cube  roots  of  8 
are  2,  and  2  a>  and  2  ufl. 

Verify  this  by  cubing  2,  2  w,  and  2  w2. 

2.    Solve:  x4  + 1  =0,  or  xA  =  -1.  (1) 

Factoring  a*+  1,        (x2  -  l)  (x2  +  t')  =  0.  (2) 

Solving  (3),  X2  =  I,  X2  =  -  i.  (5) 

Solving  (3),  X  -  ±  a/7,  X  =  ±  V— 7.  (4) 

These  are  the  -four  numbers,  each  of  which  raised  to  the  fourth  power 
equals  —  1,  or  the  four  fourth  roots  of  —  I. 

WRITTEN    EXERCISES 

1.  Find  the  3  cube  roots  of  —  1  by  solving  Xs  + 1  =  0. 

2.  Find  the  4  fourth  roots  of  1  by  solving  x*  —  1  =  0. 

3.  Find  the  6  sixth  roots  of  1  by  solving 

(aje  - 1)  =  (ar3  -  l)(x3  + 1)  =  0. 

4.  Find  the  4  fourth  roots  of  16  by  solving  x4  —  16  =  0. 

5.  Find  the  3  cube  roots  of  8  by  solving  x3  —  8  =  0. 

6.  Show  that  the  square  of  either  irrational  cube  root  of 
- 1  is  the  negative  of  the  other  irrational  cube  root. 

7.  Show  that  the  sum  of  the  three  cube  roots  of  unity  is 
zero ;  also  that  the  sum  of  the  six  sixth  roots  of  unity  is  zero. 

SUMMARY 
I.   Forms  and  Definitions. 

1.  A  general  form  of  the  quadratic  equation  is : 

ax2  +  bx  +  c  =  0.  Sec.  148. 

2.  The  general  forms  of  the  roots  of  this  equation  are : 


-6+V62- 

-4 

ac 

2a 

-&-V62- 

-4 

ac 

r2  =  — » — Y  '  Sec.  150. 

2a 


118 


ELEMENTARY   ALGEBRA 


3.  The  discriminant  of  the  quadratic  equation  is  b2  —  4  ac. 

If  5-'  _  4  ac  is  greater  than  0,  the  roots  are  real  and  unequal. 

If  },2  _  4  ac  is  equal  to  0,  the  roots  are  real  and  equal. 

If  &2  _  4  ac  is  less  than  0,  the  roots  are  complex  numbers. 

If  i,i  _  4  m  is  a  perfect  square,  the  roots  are  rational. 

If  ip.  _  4  ac  is  not  a  perfect  square,  the  roots  are  irrational. 

Sees.  157,  158. 

4.  The  discriminant  of  the  quadratic  equation  enables  us 
to  rind  by  inspection  the  nature  of  the  factors  of  a  quadratic 
expression.  Sec.  160. 

5.  In  the  quadratic  equation  sc2  +  px  +  q  =  0,  the  coefficient 
of  x  with  its  sign  changed  is  the  sum  of  the  roots,  and  the 
absolute  term  is  their  product.  Sec.  153. 

6.  Equations  of  the  form  x"  ±  a  =  0  are  binomial  equations. 

Sec.  166. 

7.  The  three  cube  roots  of  unity  are  1,  w,  and  w2,  where 

iV3 


-l  +  <-V3or_-l 


Sec.  166. 


2  2 

1 1.    Processes. 

1.  Methods  of  solution:    By  factoring,   by  completing  the 
square,  by  formula.  Sec.  152. 

2.  Certain  higher  equations  may  be  solved  by  the  methods 
of  quadratic  equations.  Sec.  16o. 

REVIEW 
WRITTEN    EXERCISES 


Solve : 

1.  x-  =  6  x  —  5. 

2.  to-  —  w  —  1  =  0. 

3.  a-2  _  6  x  -  7  =  0. 

4.  v-  +  2  v  +  6  =  0. 

5.  x2  -  5  x  +  1  =  0. 

6.  2  ar  -  .t  +  3  =  0. 

7.  3  .r  -  x  +  7  =  0. 

8.  x2  -  5x  +  11  =  0. 


9.  a2  —  14  a>  +  5  =  0. 

10.  3x*-9x-%  =  0. 

H.  a;2  -4a?-  9021  =  0. 

12.  a-2  -  4  a  +  9021  =  0. 

13.  a2  +  4  a  -  9021  =  0. 

14.  a?  +  4  .r  +  9021  =  0. 

15.  a2  +  30  .r +  221  =0. 

16.  x2  -  30  x  -  221  =  0. 


17. 

9                 9 

X        b 

18. 

ex'-  +  foe  +  a  =  0. 

19. 

(«-2)(aj  +  3)  =16. 

£0. 

x2  =  6  x  +  16. 

21. 

24  -lO.r  =  .T2. 

22. 

(x  -  1)-  =  a;  +  2. 

23. 

5  a;  +  x-2  +  6=0. 

24. 

as»  _  9  x  + 14  =  0. 

25. 

a.2  +  3  a,  _  70  _  o. 

26. 

4s2-4a;-3  =  0. 

27. 

3a2-7<c  +  2  =  0. 

28. 

or  -  10  cc  +  21  =  0. 

QUADRATIC    EQUATIONS  119 

29.  a2  -  10  x  +  24  =  0. 

30.  x2  +  10  x  +  24  =  0. 

31.  a,-2  =  9  x2  -  (a-  +  l)2. 

32.  9  x2  +  4  x  -  93  =  0. 

33.  4  x2  +  3  x-  -  22  =  0. 

34.  6  x2  -  13  a?  +  6  =  0. 

35.  (x+  l)(a?4-2)  =  x  +  3. 

36.  — -      -  +  - =  c. 

O  +  X       0  —  x 

37.  — *—       —^  =  aa-  +  62- 
a  -|-a; 

38     8  +  H         «  +  12     =1 
a; +  2       2(a?  +  19)     2* 


i)^2,+3^-2). 


40 


x-3  2 

«  —  a      a;  —  &  _        (a  —  Iff 


x  —  b      x  —  a      (x  —  a)(x—b) 

41.  (as  -  l)\x  +  3)  =  *(*  +  5) (a;  -  2). 

42.  x2  -  6  aca-  +  a2(9  c2  -  4  b2)  =  0. 

State  what  can  be  known  by  means  of  the  discriminant 
and  without  solving,  concerning  the  factors  of  the  following 
trinomials  : 

43.  3x2-2x  +  l.  45.    5 v2  +  20 y  +  20. 

44.  4ar  +  lla;-l.  46.    2x2-x  +  3. 

Similarly,  what  can  be  known  about  the  roots  of : 

47.  *2-9  =  0?  49.   z2-*-l  =  0? 

48.  5x2  +  a,-  +  2  =  0?  50.   5a2  =  9a;? 

How  must  a  be  chosen  in  order  that : 

51.  The  roots  of  x2  +  ax  -f-  5  =  0  shall  be  imaginary  ? 

52.  The  roots  of  ax2  +  6  x  +  1  =  0  shall  be  real  ? 


120  ELEMENTARY   ALGEBRA 

53.  The  roots  of  x2  +  4  x  +  2  a  =  0  shall  be  real  and  of  oppo- 
site signs  ? 

54.  The  roots  of  (a  -f  1)  x2  +  3  x  —  2  =  0  shall  be  imaginary  ? 

55.  The  roots  of  -Ix2  —  ax +  2  =  0  shall  be  real  and  both 
positive  ? 

Find  the  values  of  ra  for  which  the  roots  of  the  following 
equations  are  equal  to  each  other.  What  are  the  correspond- 
ing values  of  x? 

56.  x2  —  12aj  +  3m  =  0.        .     58.   4x2  +  mx  +  x  +  1  =  0. 

57.  mx2  +  8  x  +  m  =  0.  59.    mx2  -f  3  mx  —  5  =  0. 

60.  A  number  increased  by  30  is  12  less  than  its  square. 
Find  the  number. 

61.  The  product  of  two  consecutive  odd  numbers  is  99. 
What  are  the  numbers  ?     Is  there  more  than  one  set  ? 

62.  Find  two  consecutive  even  numbers  the  sum  of  whose 
squares  is  164. 

63.  Find  a  positive  fraction  such  that  its  square  added 
to  the  fraction  itself  makes  f . 

64.  If  a  denotes  the  area  of  a  rectangle  and  p  its  perimeter, 
show  that  the  lengths  of  the  sides  are  the  roots  of  the  equation 

x2  —  -zX-\-a  =  0. 

65.  The  diagonal  and  the  longer  side  of  a  rectangle  are  to- 
gether equal  to  5  times  the  shorter  side,  and  the  longer  side 
exceeds  the  shorter  by  35  meters.  Find  the  area  of  the  rec- 
tangle. 

66.  A  rug  9  ft.  by  12  ft.  covers  \  the  floor  of  a  room,  and 
can  be  laid  so  that  the  uncovered  strip  of  floor  about  the  rug 
is  of  the  same  breadth  throughout.  Find  the  dimensions  of  the 
room. 

67.  A  company  of  soldiers  attempts  to  form  in  a  solid 
square,  and  56  are  left  over.  They  attempt  to  form  in  a 
square  with  3  more  on  each  side,  and  there  are  25  too  few. 
How  many  soldiers  are  there  ? 

68.  It  took  a  number  of  men  as  many  days  to  dig  a  ditch  as 


QUADRATIC   EQUATIONS 


121 


there  were  men.     If  there  had  been  G   more  men  the   work 
would  have  been  done  in  8  days.     How  many  men  were  there  ? 

69.  Solve:  2x2  +  Qx  +  c  =  0. 

What  value  has  c  if  the  two  values  of  x  be  equal  ?    If  they  be 
reciprocal  ? 

70.  A  public  library  spends  $180  monthly  for  books.  In 
June  the  average  cost  per  book  was  15  ^  less  than  in  May,  and 
60  books  more  were  bought.  How  many  books  were  bought 
in  May  ? 

71.  When  water  flows  from  an  orifice  in  a  tank  the  square 
of  the  velocity  (v)  equals  2  g  times  the  height 
(h)  of  the  surface  above  the  orifice.  Write 
the  equation  that  denotes  this  fact,  g  is 
the  "  constant  of  gravity  "  and  may  be  taken 
as  32. 

72.  What  does  the  square  of  the  velocity, 
(v2),  at  A  in  Fig.  1  equal  ?  Find  this  velocity.  Fig.  l. 

73.  What  would  be  the  velocity  of  the  water  if  an  opening 
were  made  halfway  up  the  tank  shown  in  the  figure  ? 

74.  Find  the  velocity  with  which  water  rushes  through  an 
opening  at  the  base  of  a  dam  against  which  the  water  stands 
25  ft.  high. 

75.  The  distance  on  the  level  from  the  bottom  of  the  vertical 
wall  to  the  point  where  the  stream  reaches 
the  ground  is  called  the  range.  The  range 
in  Fig.  2  is  a.  If  t  is  the  number  of  sec- 
onds taken  by  the  water  to  reach  the  ground 
after  leaving  o,  then  a  =  vt.  Take  a  to  be 
10  ft.,  and  find  t  after  computing  v  as  above. 

76.  Water  flowing  from  a  leak  in  a  dam  reaches  the  plane  of 
the  base  20  ft.  from  the  dam.  The  opening  is  10  ft.  below 
the  surface  of  the  water.  How  long  after  leaving  the  dam 
does  the  water  strike  the  base? 

77.  Water  flowing  from  an  orifice  in  a  standpipe  reaches  the 
level  ground  40  ft.  from  the  base  in  3  seconds, 
the  column  of  water  above  the  orifice  ? 

9 


K----CL--* 


Fig.  2. 


How  high  is 


122  ELEMENTARY   ALGEBRA 


SUPPLEMENTARY  WORK 
ADDITIONAL  EXERCISES 


Solve  for  x : 

1.1  1 


a      a  4-  x      a2  —  x2 


=  0. 


Ill  1 

2-    -  +  TH 7 =  0. 

a      b      x      a  -\-b-\-x 

3.  (2  a  -  5  -  a;)2  +  (3  a  -  3  a-)2  =  (a  +  5  -  2  a-)2. 

4.  (3£-4a  +  3&)2  +  (2a  +  a)2=(>-4a+&)2 

4-(2.e-3a4-4&)2. 

5.  (7a  +  3&+a)2+(4a-&-8a)2-(4a  +  3&  +  4a)2 

=  (7  a  +  &  -  6  a)2. 

6.  (2a  +  4&  +  6z)(3a-9&  +  a)  =  (2a-5&  +  5x)2. 

7.  (a;  +  a)(5  a  -  3  a  -  4  &) = (a  +  a  -  2  6)2. 

8.  (5  x  +  4  a  +  3  &)(10  aj  -  6  a  +  8  6)  =  (5  x  +  a  4-  7  6)2. 

9.  (26a;  +  a 4- 22  6)  (14  a?+ 13 a - 26)  =  (16 x  +  lla  +  S  b)2. 
10.    (8c4-10  4-4x)(18c4-160  4-24z)  =  (12c  +  40  +  lla;)2. 

14/4-16       2/4-8       2tf 


11 
12 


21  8/  -11 

a2  +  x2      (b  +  c)2 


a2  (b  -  c) 


2 


i«     3a-2-27  ,  90  +  4X2 
13-      ^  +  3    +-^-^9--7' 

.t2+x-2       ^  +  1_Q 
x*  +  2x-3     a>-3 

1  111 

15.  — -= =  -+r 

a-\-b—x     a      b      x 

,n     /         sA      3a  +  3aA      _      l-2a,       Qx       l  +  « 

16.  (a-x)^l--_rj-2  =  r-^(c-3)--_I. 

2c 


QUADRATIC   EQUATIONS  123 

Quadratic  Equations  involving  Radicals 

If   the   equation   involves   but   one  radical,  the  method  of 
Sec.  96,  p.  48,  can  be  used. 

When  the  equation  resulting  from  squaring  is  not  directly 
solvable,  substitution  is  sometimes  useful, 

EXAMPLE 


Solve:        5x?-3x  +  ^5xi-3x  +  2  =  18.  (1) 


Putting  5x2-3x=jr, 
the   given   equation  be- 
comes 


y  +  Vy  +  2  =  18.  (2) 

Subtracting  y,                                         \/y  +  2  =  18  —  y.  (3) 

Squaring,                                                y  +  2  =  324  -  36  y  +  y2.  (4) 

Rearranging,                             y2  —  37  y  +  322  =  0.            (5) 

»      37±V372-4-322  ^ 

Solving,                                                                y  =  — (6) 

=  23  or  14.  (7) 

=  23,  (*) 

and                                                    5x2-3x  =  14.  (9) 

Solving  these,                                                        X  =  — — ,  (10) 

and                                                                    X  —  2.  or  —  1.4.  (11) 
Test.     Substituting,  it  appears  that  2  and  —  1.4  satisfy  the  equation. 

The  values  °  ±  ^         satisfy 
10 


Hence   the  values   of  x  0 

are  determined  from  OX"  —  o  X  —  Z.3,  (A; 


6x2  -  3x  -  V5x2  -  3x  +  2  =  18. 
Sometimes  successive  squaring  is  necessary. 

EXAMPLE 


Solve:  V2x  +  6+V3x  +  l=8.  (1) 

Rearranging,  V2x  +  6  =  8  -  V3x  +  1.  (£) 

Squaring,  2  x  +  6  =  64-16  V3x-fl  +3  x+1.  (3) 

Collecting,  16V3X+1  =  X  +  59.  (4) 

Sqconectinggain  aDd  x2  - 650  X  +  3225  =  0.  (5) 

Solving,  x  =  5,  or  645.  (6) 

Test.  Trial  shows  that  the  first  of  these  values  satisfies  the  given  equa- 
tion ;  and  it  is  obvious  on  inspection  that  the  second  cannot  satisfy  the 
equation. 


124  ELEMENTARY   ALGEBRA 

Sometimes    it    is  best   first   to    transform    the    given    ex- 
pression. 

EXAMPLE 


Solve:  2x  +  VI¥+9  =  -2x  +  1+^5x  +  6 

2  x  -  V4  ar  +  9 


CO 


Clearing  of 


fractions,  0  =  2  X  +  1  +  V5  X  +  6.  (2) 

Rearranging,  8  —  2  X  —  y/b  X  +  6.  (3) 

Squaring  and 
collecting,         4  Xz  -  37  £  +  58  =  0.  (4) 

Hence,  x  =  2,  or  7J.  (5) 

Test.  By  trial,  2  is  seen  to  satisfy  the  given  equation.  To  avoid  the 
complete  work  of  substituting  7^,  we  note  that  every  root  of  (2)  must 
satisfy  (3).  If  x  =  7 J,  the  left  member  of  (3)  is  negative  and  the  right 
member  positive.  Hence  7^  is  not  a  root.  It  would  satisfy  (5)  if  the 
radical  had  the  negative  sign. 

WRITTEN    EXERCISES 
Solve : 


1.  a-V3o;  +  10  =  6.  3.    v^r-Vaj-15  =  l. 

2.  6 x- «*- 12  =  0.  4.   x%-x$-6  =  0. 


5.  x2  +  x  +  VaF+  x  +  1  =  —  1. 

6.  x-  +  a  +  V#2  +  2  a  =  —  a. 


7.  x-2 _j_ 6 a _ 9  =  Vx2  +  6a  +  7. 

8.  4  Var  +  7  a;  +  10  =  Vo  •  Var9  +  5  a;  +  10. 
9.    Var'  +  6  a;  -  16  +  (x  +  3)2  =  25. 

10.  2*-5»*  +  2  =  0.  is.    V^W5  +  — 4==5- 

11.  af*  — af*— 6=0.  Va?— 5 

12.  3  x~%  -  4  a?"*  =  7.  14.    a; +  2  + (a? +  2)*  =  20. 


15.  ar5  +  3.x-3  +  Va;2  +  3a;+17  =  0. 

16.  Va7+7+V3a?-2  =  -^±^-* 

V3a;-2 

17.  ,,;  +  y,^T8  =  3a;-V^-^. 

a;  —  Var*  —  8 


18.    Vx(a  +  .r)  +  Vx(a  —  x)  =  2 Vax. 


QUADRATIC   EQUATIONS  125 

Factoring  applied  to  solving  Equations 
Factor  Theorem.     The  relation 

y?  +px  +  q  =  (a  —  n)  0  — »») 

is  a  special  case  of  an  important  general  theorem,  known  as 
the  factor  theorem : 

If  any  polynomial  in  x  assumes  the  value  zero  when  a  is  sub- 
stituted for  x,  then  x  —  a  «s  a  factor  of  the  polynomial. 

We  prove  this  first  for  the  particular  polynomial 

x*  -  3  x3  +  7  x2  -  2  x  -  10. 

By  substituting  2  for  x  we  find  that  the  polynomial  assumes  the  value 
zero. 

Suppose  the  polynomial  to  be  divided  by  x  —  2,  and  denote  the  quo- 
tient by  Q  and  the  remainder  by  B,  the  latter  being  numerical. 

Then  x*  -  3  xs  +  7  x2  -  2  x  -  16  =  (x  -  2)  §  +  B. 

In  this  equation  substitute  2  for  x ;  the  left  number  becomes  zero  as 
just  seen ;  2  —  2  is  also  0,  and  0  times  Q  is  0,  no  matter  what  value  Q 
may  have  ;  hence  the  result  of  substitution  is, 

0  =  0  +  B,  or  B  =  0. 

Consequently  x4  -  3  x3  +  7  x2  —  2  x  —  10  =  (x  —  2)  Q,  or  x  —  2  is  a 
factor  of  x4  —  3  x3  +  7  x2  —  2  x  —  16.  This  may  now  be  verified  by  actual 
division,  but  the  value  of  the  theorem  lies  in  the  fact  that  by  it  we  know 
without  actual  division  that  x  —  2  is  a  factor  of  the  polynomial. 

Notation  for  Polynomials.  It  is  convenient  to  have  a  symbol  to  repre- 
sent any  polynomial  involving  x.  For  this  purpose  the  symbol  P(x) 
is  used.  It  is  read  "  P  of  x."  P(2)  means  the  same  polynomial  when  2 
is  substituted  for  x,  P(—  V5)  means  the  polynomial  when  —  V5  is  put  for 
x  ;  and  P(a)  means  the  polynomial  when  a  is  put  for  x. 

General  Proof:  Let  P(x)  denote  the  given  polynomial. 

Suppose  P(x)  to  be  divided  by  x  —  a.  There  will  be  a  certain  quotient, 
call  it  Q(x),  and  a  remainder,  B.     This  remainder  will  not  involve  x, 

otherwise  the  division  could  be  continued. 

» 

We  have  then :  P(x)  =  (x  -  a)  Q  (x)  +  B.  (1) 

In  this  equation,  put  a  in  place  of  x : 

P(a)  =  (a-a)Q(a)  +  B.  (8) 


12G  ELEMENTARY   ALGEBRA 

By  hypothesis,  P(a)  =  0,  a  —  a  =  0,  and  R  remains  unchanged  when 
x  is  replaced  by  a,  for  there  is  no  x  in  R.     Hence 

0  =  0  +  R,  or  R  =  0.  (3) 

Substituting  this  value  of  R  in  (1), 

P(x)  =  (x-a)Q(x).  (4) 

That  is,  x  —  a  is  a  factor  of  P  (x) ,  as  was  to  be  proved. 

WRITTEN     EXERCISES 

In  each  polynomial  substitute  the  values  given  for  x,  and 
use  the  factor  theorem,  when  applicable,  to  determine  a  factor 
of  the  polynomial : 

Polynomial  Values  fok  » 

1.  x*  +  3  x3  +  4  x2  -  12  x  -  32  1,  2,  -  2. 

2.  x*  -  7  x3  + 14  x2  +  x  -  21  1,  -  1,  2,  3. 

3.  2x4  +  5x3-41x2-64x  +  80  4,-4,5,-5. 

By  use  of  the  factor  theorem,  prove  that  each  polynomial 
has  the  factor  named: 

Polynomial  Factor 

4.  x3  +  12x2  +  31  x- 20  x  +  5. 

c     a^-l      a8-l 

5. x—  a. 

x  —  1       a—  1 

6.  ^_2x-2  +  — -fm3-2m2  +  — ^  a-m. 

m      \  x  y 

7.  (x-a)2  +  (x-^)2-(a-6)2  x-Z>. 

8.  x3  +  2x2  +  3x  +  2  x  +  1. 

9.  x3"  +  2x2n  +  3x"  +  2  xn  +  l. 

10.  x3  +  ax2  —  aV  —  a8  x  — a2. 

11.  9x5+(3a2-12a)x4-4a3x3+3a2x+a4       x  + 


a2 
3' 


Roots  and  Factors.  The  factor  theorem  may  be  worded 
thus : 

If  a  is  a  root  of  the  equation  P(x)  =  0,  then  x  -  a  f s  a  factor 
of  the  polynomial  P(x). 


QUADRATIC    EQUATIONS  127 

It  is  also  evident  that : 

If  x  —  a  is  a  factor  of  P(x),  then  a  is  a  root  o/P(x)  =  0. 

For,  let  P(x)  =  (x—  a)  Q  (*). 

Putting  a  for  x,  P(a)  =  (a  —  a)(j)(rt). 

=  0. 

Thus  there  is  a  complete  correspondence  between  root  of 
equation  and  linear  factor  of  'polynomial.  As  in  quadratic 
polynomials,  so  also  in  all  higher  polynomials :  To  every  linear 
factor  there  corresponds  a  root,  and  vice  versa.  If  we  know  a 
linear  factor,  we  can  at  once  determine  a  root ;  if  we  know  a 
root,  we  can  at  once  write  a  linear  factor. 

This  property  enables  us  to  write  out  any  equation  of  which 
all  the  roots  are  known. 

EXAMPLE 

Write  the  equation  whose  roots  are  2,  3,  —  1,  0. 

The  factors  are  x  -  2,  x  —  3,  x  —  (—  1),  and,  x  —  0. 
The  polynomial  is  (x  —  2) (x  —  3)  (x  +  l)x. 
The  equation  is  x4  —  4  x3  +  x2  +  6  x  —  0. 

WRITTEN    EXERCISES 

Write  the  equations  whose  roots  are : 

1.  3,  —  4,  0.  5.    a,  b,  c. 

2.  2,  —  2,  V5,  —  V5.  6.    a,  —  b,  c. 

3.  4,  1  +  V2,  1  -  V2.  7.    a,  -  a,  b,  -  b,  0. 

4.  6,  - 1,  i,  -  i.  8.   2  -  V3  i,  2  +  V3  i,  4. 

When  a  root  is  known,  we  thereby  know  a  factor  of  the 
given  polynomial,  and  this  may  aid  in  finding  the  other  roots 
of  the  equation. 

EXAMPLE 

Find  the  roots  of  the  equation  x3  —  3  xl  —  4  x  + 12  =  0. 

The  equation  x3  -  3x2 -4x+ 12  =0  (i) 

has  the  root  —  2.  (Verify  by  substitution.)  Hence  the  polynomial  has 
the  factor  x  —  (—  2)  or  x  +  2.  Finding  the  other  factor  by  division,  the 
equation  may  be  written, 

(x  +  2)(x2-5x  +  6)  =  0.  (JS) 


128  ELEMENTARY   ALGEBRA 

The  left  member  will  be  zero,  either  if 

x  +  2  =  0,  (3) 

or  x2-5x  +  G  =  0.  (4) 

Equation  (3)  has  the  root  —  2,  and  equation  (4)  has  the  roots  2  and  3. 
Hence  the  roots  of  (2)  are  2,  —  2,  and  3. 

WRITTEN    EXERCISES 

Each  of  the  following  equations  has  oue  of  the  six  numbers, 
±  1,  ±  2,  ±  3,  as  a  root.  Find  one  root  by  trial,  and  then  solve 
completely  as  in  the  example  : 

1.  xs  +  5x2  —  4:x-20  =  0.  5.   ar3  +  a2  +  a-  +  l  =  0. 

2.  xs  +  3x2-16x-±8  =  0.  6.    x9  +  3x2-5x-10  =  0. 

3.  ^-3^  +  5^-15  =  0.  7.    2a3-2a2-17a  +  15  =  0. 

4.  xs  +  2x2-ax-2a  =  0.  8.    a-3 +  3  a-2  -4a--  12  =  0. 

9.    x3  -  (2  b  +  3)  x2  +  (b2  +  6  &  -  c)  x  +  3  &2  -  3  c2  =  0. 

10.  By  finding  the  h.  c.  f.  of  their  first  members,  find  the 
two  roots  that  are  common  to  the  equations 

aj*_4o3  +  2<B2  + a;  +  6  =  0  and  2a-3-9a-2  + 7a-  +  6  =  0. 

Whenever  an  equation  has  the  right  member  zero,  any 
factors  that  may  be  given  are  aids  to  the  solution.  The  roots 
corresponding  to  factors  of  the  first  degree  can  at  once  be 
written  out.  Factors  of  higher  degree  determine  equations  of 
corresponding  degree  to  be  solved. 

WRITTEN    EXERCISES 
Solve  by  equating  each  factor  to  zero : 

1.  a(a--l)(a-  +  2)  =  0.  4.    x(x  +  2i)(x-7 V§)  =  0. 

2.  (a  +  2)(a--5)(a  +  V3)  =  0.     5.    («  -  4)  (a8  -  4)  (x  +  4)  =  0. 

3.  (a-2  -l)(x  -  7)(x  +  5)  =  0.       6.    x(x  -  ai)(x*  -  62)  =  0. 

7.  x  (x  -  7) (x  +  V7)(a-°  +  7)  =  0. 

8.  (a-2  +  5.T  +  6)(a2-9a-  +  14)  =  0. 

9.  (a2-7a-  +  ll)(.r  +  2a  +  5)  =  0. 


QUADRATIC   EQUATIONS 


129 


10.  x(x2-4)(x2  +  7x-3)=0. 

11.  (x2-  25)  (a?  +  3)  (a?  -  6  x  + 13)  =  0. 

12.  (x  —  a)  (x  +  6)  .r  (ar  —  2  ex  +  c2)  =  0. 

13.  (ar  +  2  ox-  +  6)  (x2  -  a2b)  (5  x  - 1)  =  0. 


Graphical  Work 

Graphical  Representation  of  Solutions.  The  solutions  of  ax2  + 
hx  -f-  c  =  0  may  be  represented  graphically  by  plotting  the  curve 
corresponding  to  y  =  ax2  -+-  bx  +  c.  The  solutions  of  the  equa- 
tion will  be  represented  by  the  point  where  the  curve  cuts  the 
iK-axis. 

EXAMPLE 

Represent  the  solutions  of  x2  -f-  x  —  2  =  0. 

To  draw  the  curve  y  —  x2  + 
x  —  2,  we  make  a  table  of  a 
sufficient  number  of  pairs  of 
values  of  x  and  y,  plot  the  cor- 
responding points,  and  sketch  a 
smooth  curve  through  them. 


-4 

10 

-3 

4 

-2 

0 

-1 

-2 

3 

-2& 

1 

2 

-21 

1 

4 

_  9  3 

0 

-2 

1 

0 

2 

4 

3 

10 

4 

18 

Note.  When  the  integral 
values  do  not  satisfactorily  out- 
line the  shape  of  the  curve,  frac- 
tional values  of  x  must  also  be 
used  as  above. 


130  ELEMENTARY   ALGEBRA 

This  graph  not  only  represents  the  solutions  of  ar  +  x— 2  =  0; 
but,  what  is  far  more  important,  it  represents  the  values  and 
variation  of  the  polynomial  x2  +  x  —  2  for  all  values  of  x 
(within  the  limits  of  the  figure).  Thus  EF  represents  the 
value  of  the  trinomial  for  x  =  \,  and  CD  that  for  x=—  2\. 
The  figure  also  tells  us  that  x2  +  x  —  2  is  positive  when  x  is 
negative  and  numerically  greater  than— 2  (algebraically  less 
than  —  2).  That  when  x  is  algebraically  less  than  —2  and  in- 
creasing, the  trinomial  is  positive  and  decreasing,  reaching 
the  value  zero  when  x=—2.  As  x  increases,  the  trinomial 
continues  to  decrease,  becoming  •  negative,  reaching  its  least 
value  for  x  =  —  \.  It  then  begins  to  increase,  through  negative 
values,  reaching  zero  when  x  =  l.  As  x  increases  further,  the 
trinomial  continues  to  increase,  becoming  positive  and  continu- 
ing to  increase  as  long  as  x  increases. 

The  graph  enables  us  to  read  off  approximately  not  only  the 
roots  of  x2  +  x  —  2  =  0,  but  also  those  of  any  equation  of  the 
form  x2  +  x  —  2  =  a,  if  they  are  real. 

For  example : 

1.  The  roots  of  x2  +  x  —  2  =  3  are  indicated  by  the  points  where  the 
line  drawn  parallel  to  the  x-axis  through  the  point  3  on  the  y-axis  cuts  the 
graph. 

2.  The  line  parallel  to  the  x-axis  through  the  point  —  3  does  not  cut 
the  graph  at  all.  This  tells  us  that  there  are  no  real  values  of  x  which 
make  the  trinomial  x-  +  x  —  2  equal  to  —3.  The  equation  x'2+x— 2 
=  —  3  has  imaginary  roots. 

The  graph  shows  (1)  that  the  equation  x2  +  x  —  2  =  a  has  two  real  and 
distinct  roots  whenever  a  is  positive ;  also  when  a  ranges  from  zero  to  a 
little  beyond  —  2  ;  (2)  that  it  has  imaginary  roots  when  a  is  less  than  a 
certain  value  between  —  2  and  —  3  ;  (3)  at  a  certain  point  the  line  paral- 
lel to  the  x-axis  just  touches  the  graph.  This  corresponds  to  the  case  of 
equal  roots  of  the  equation.  The  value  of  a  in  this  case  may  be  read 
from  the  graph  as  —  2\.  This  means  that  the  equation  x2  +  x  —  2  =  —  2J, 
or  x'2  +  x  +  \  =  0,  has  equal  roots.  This  may  be  verified  by  solving  the 
equation  or  by  applying  the  discriminant  (Sec.  384,  p.  349). 

Note.  Since  even  the  best  of  drawing  is  not  mathematically  accurate, 
results  read  off  from  a  graph  are  usually  only  approximately  correct. 
The  closeness  of  the  approximation  depends  on  the  degree  of  accuracy  in 
the  drawing. 


QUADRATIC    EQUATIONS  131 

WRITTEN    EXERCISES 
Treat  each  trinomial  of  Nos.  1  to  9  below  as  follows : 

(1)  Draw  the  graph  of  the  trinomial. 

(2)  From  the  graph  discuss  the  variations  of  the  trinomial. 

(3)  From  the  graph  read  off  approximately  the  roots  of  the 
equation  resulting  from  equating  the  trinomial  to  zero. 

(4)  If  the  trinomial  be  equated  to  a,  read  from  the  graph 
the  range  of  values  of  a,  for  which  the  roots  of  the  equation 
are,  (i)  real  and  distinct,  (ii)  real  and  equal,  (iii)  imaginary. 

(5)  Last  of  all,  verify  those  of  the  preceding  results  which 
relate  to  roots  of  the  equations  by  solving  the  equations  or  by 
use  of  the  discriminant. 

x2  +  4  x  —  5.  7.   x2  +  2  x  +  3. 

3a2-  2x.       .        8.    x*-2x-l. 

ar-4.  9.    8  +  2X-X2. 

10.  Draw  the  graph  of  the  function  2  x*  +  5&  — 3,  and 
state  how  it  varies  as  x  varies  from  a  negative  value,  numer- 
ically large  at  will,  through  zero  to  a  large  positive  value.  For 
what  values  of  x  is  the  function  positive  ?  For  what  values 
negative  ? 

11.  What  is  the  graphic  condition  that  ax2  +  bx  +  c  shall 
have  the  same  sign  for  all  values  of  x  ?  What  must  therefore 
be  the  character  of  the  roots  of  ax2  +  bx  +  c  =  0,  if  the  tri- 
nomial ax*-\-bx  +  c  has  the  same  sign  for  all  values  of  x? 

Determine  m  so  that  each  of  the  following  trinomials  shall 
be  positive  for  all  values  of  x : 

12.  x2  +  mx  +  5.       13.  3X2  —  5x  +  m.       14.   mx2  +  6  x  +  8. 


1.   x2  +  5x  —  6. 

4 

2.   x*  +  3x  +  2. 

5 

3.   2a?  —  5aj  +  2. 

6 

CHAPTER   VIII 
SYSTEMS  OF  QUADRATIC  AND  HIGHER  EQUATIONS 

SIMULTANEOUS    QUADRATIC    EQUATIONS 

167.  Two  simultaneous  quadratic  equations  with  two  un- 
knowns cannot  in  general  be  solved  by  the  methods  used  in 
solving  quadratic  equations,  because  an  equation  of  higher 
degree  usually  results  from  eliminating  one  of  the  unknoAvns. 
But  many  simultaneous  quadratic  equations  can  be  solved  by 
quadratic  methods,  and  certain  classes  of  these  will  be  taken 
up  in  this  chapter. 

168.  Class  I. 

A  system  such  that  substituting  from  one  equation  into  the  other 
produces  a  quadratic  equation. 

1.  A  system  of  equations  composed  of  a  linear  equation  and  a 
quadratic  equation  can  always  be  solved  by  substitution. 


EXAMPLE 

J3a;  +  42/  =  24. 

CO 

re : 

{     x>  +  y*  =  25. 

(2) 

From  (i), 

3  x  -  24  -  4  y. 

(3) 

From  (3), 

x  =  8 ^« 

3 

(4) 

Substituting  (4)  in  (2), 

(8-^)2+2/2  =  25. 

{5) 

Simplifying  (5), 

25  2/2-192?/  +  351  =0. 

(G) 

Factoring  (6), 

O/-3)(25.y-ll7)=0. 

(7) 

Solving  (7), 

y  =  3  and  2/  =  —  • 
9                 J        26 

(8) 

From  (/,), 

44 

x  =  4  and  x  = 

25 

(9) 

132 


SIMULTANEOUS   QUADRATIC   EQUATIONS         133 

Test.      3  . 4  +  4  .  3  =  24  and  ?Li*  +  i^l  =  24. 

25  25 

32  +  42  _  25. 

It  should  be  noted  that  to  every  x  there  corresponds  only  one  particular 
value  of  y,  and  vice  versa.  The  proper  correspondence  may  be  seen  by 
noticing  which  value  of  one  unknown  furnishes  a  given  value  of  the  other 
in  the  process  of  solution.     In  the  example  above,  y  =  3  produces  x  =  4. 

2.  When  both  equations  are  quadratic,  substitution  is  appli- 
cable if  the  result  of  substitution  is  an  equation  having  the  quad- 
ratic form. 

EXAMPLE 

•      (x2  +  y2  =  25.  (1) 

1        ay  =  12.  {2) 


Solve 


12 

From  (2),  X  = (3) 

y 

(10\2 
—  j    +  y2  —  25.  (4) 

Simplifying  (U),  we  obtain  an  „4  _  05  u2    1    1 44  _  n  (k\ 

equation  in  quadratic  form,  J  y    ~  •  \   J 

Factoring  (5),  (y*  -  16)  (^  -  9)  =  0.  (6) 

Solving  (6),  y  =  ±  4,  and  ±3.       (7) 

Substituting  (7)  in  (3),  X  =  ±  3,  and  ±  4.        (5) 

Test.     Taking  both  values  to  be  positive,  or  both  to  be  negative, 

(±3)2  +  (±4)2  =  25. 

(±3)  -±4  =  12. 

There  are  frequently  various  methods  of  solving  the  same  problem. 
Thus,  in  the  last  example,  multiply  (#)  by  2,  add  it  to  or  subtract  it  from 
(1)  ;  the  resulting  equations  (x  +  y)'2  —  49,  and  {x  —  y)2  =  9,  can  be 
solved  by  extracting  the  square  roots,  and  adding  and  subtracting  the 
results. 

„   ,  _  WRITTEN    EXERCISES 

Solve  and  test : 

1.  x2  +  y*  =  13,  3.  3  x2  -  2  xy  =  15, 
2x  +  3y  =  13.  2x+3y  =  12. 

2.  2x2-y2  =  U,  4.  x2-xy  +  y2  =  3, 
3x  +  y  =  ll.  2x  +  3y  =  8. 


134  ELEMENTARY    ALGEBRA 

5.  x2  +  y2  =  2xy,  11.  x  —  y  =  l, 

x  +  y  =  8.  x2  —  y2  =  16. 

6.  ar  +  y2  =  25,  12.  3xy  +  2x  +  y  =  4S5, 
x  +  y  =  l.  3x-2y  =  0. 

7.  x  —  y  =  1£-,  13.  .r -}->/  =  a, 
a#  =  20.  .c2  +  ?/2  =  &. 

8.  z2-?/2  =  16,  14.  x  —  y  =  b, 
x-y  =  2.  xy  =  a2. 

9  ?L=^_i^  15-  2x  =  x2-y2> 

'  y2  ~  9        ?/  '  2  a  =  4  xy. 

x-y  =  2.  ±      4 

16.   -=P-, 

10.  2z  +  ?/=7,  2/     ooj 

ar  +  2?/2  =  22.  z2-?/2  =  81. 

169.    Class  II. 

A  system  in  which  one  equation  has  only  terms  of  the  second 
degree  in  x  and  y  can  be  solved  by  finding  x  in  terms  of  y,  or 
vice  versa,  from  this  equation  and  substituting  in  the  other. 


Solve 


( x2  —  5  xy  +  6  y2  =  0. 

CO 

re: 

a;2-/ =  27. 

(*) 

Dividing  (J)  by  y2, 

x2     Say     °y2  =  o, 
J/2       2/2        y2 

(5) 

Simplifying  (3), 

(iM;)+6=0- 

(4) 

Factoring  (4), 

(H(H=ft 

(5) 

Solving  (5), 

?  =  2,  -  =  3. 

2/       y 

(e) 

From  (6), 

x  =  2  y,  x  =  3  y. 

(7) 

Substituting  x  =  1y  in  (2), 

(2  y)2  -  ,/-•  =  27. 

(*) 

Solving  (*■), 

y=±3. 

(0) 

SIMULTANEOUS   QUADRATIC   EQUATIONS         135 


x  =  ±6. 


Substituting  (9)  in  x=2y, 
Substituting  se=  3  y  in  (2), 

Solving  (11), 

Substituting  (12)  in  x=Zyl 


Test. 

Taking  both  values  to  f  (±6)2-5(±  3)  (±6)  +  6(±  3)2  =  0. 

be  positive  or  both  -{ 

negative,  I  (±6)2  -  (±3)2  =  27. 


(10) 
(11) 
(12) 

(13) 


Taking  the    signs   as 
before, 


±  9  \/6\2 


-^m)+°m=°- 


{m°-m=*- 


Notes.     1.  When  the  equation  in  -,  as  in  the  fourth  step,  cannot  he 

y 

factored  hy  inspection,  the  formula  for  solving  the  quadratic  equation  is 
used.     Putting  -  —  s,   a  quadratic  equation  is  obtained  whose   solution 

y 

will  lead  to  the  results  of  step  (7)  above. 

2.  When  an  equation  has  the  right  member  zero,  it  is  unnecessary  to 
divide  by  x2  or  y2,  if  the  left  member  can  be  factored  by  inspection. 
Thus,  in  (1)  above,  (x  —  3  y)  (x  —  2  y)  =  0,  hence  x  =  3  y  and  x  =  2  y,  as 

in  (7). 


Solve : 


WRITTEN    EXERCISES 


1.  x2-2xy-3y°-  =  0, 
xr  +  2y2  =  12. 

2.  a?  +  Xy  +  tf  =  0, 

^  +  ^  =  -1. 

3.  6  x2  4-  5  xy  +  y2  =  0, 

y*-.X-y  =  32. 

4.  2a?  —  5xy  +  2y2  =  0, 
4aj8-4/  +  3«  =  330. 

5.  x(x  +  ?/)  =  0, 

a2  _  ^  +  y-  =  27. 


6.    x2  —  ?/2  =  0, 


ar6 


3xy  +  y  =  105. 


7.  3x2-2xy-?/2  =  0, 

8.  4  or  +  4  £?/  +  ,?/'-'  =  0, 
x  +  3y2-2x  =  195. 

9.  x2  +  3a;-4  ?/  +  xy  =  33, 
x2  +  7  x?/  + 10  if  =  0. 

10.    iB24-aV  =  0, 

a2  +  ?/-/  =  - a2. 


136  ELEMENTARY   ALGEBRA 

170.      Class  III. 

A  system  of  two  simultaneous  quadratic  equations  whose  terms 
are  of  the  second  degree  in  x  and  y  ivith  the  exception  of  the  ab- 
solute terms  can  be  solved  by  reducing  the  system  to  one  of 
Class  II. 

This  can  be  done  in  two  ways : 

1.  Make  their  absolute  terms  alike,  and  subtract.  The  result- 
ing equation  has  every  term  of  the  second  degree  in  x  and  y. 

EXAMPLE 

Solve:  f^  +  ^  =  66,  (i) 

\  rf  -  f  =  11.  (2) 

Multiplying  (2)  by  6,                       6  X2  —  6  y2  =  66.  (3) 

Subtracting  (i)  from  (3),      5  X2  —  xy  —  6  y2  —  0.  (4) 

Factoring  (4),                     (5  X  —  6  y)  (X  +  y)  =  0.  (5) 

Expressing  x  in  terms  of  y,       X  —  §  y  and  X  =  —  y.  (6) 

Substituting  as=§ y  in  (2),                §|  y2  —  y2  —  11.  (7) 

Solving  (7),                                                      ?/  =  ±  5.  (5) 

Substituting  i/=±  5  in  a;  =  |  y,                          X  =  ±  6.  (9) 

Similarly,  substituting  as=-  y  in (2),  y2  —  y2  =  11.  (10) 

But  this  leads  to                                                 0  =  11.  (11) 

(Zi)  being  impossible,  the  solution  is 

x  =  ±  6,  y  =  ±  6.  (12) 

TEST.      Taking    the    values  [  (±6)2  +  (  ±6)(  ± 5)  =  66. 
to  be  both  positive  <  f +6Y2— f  +  5V2  =  11 

or  both  negative,      {  \^    /        \  ^    / 

2.  Substitute  vxfor  y  throughout  the  equations  and  solve  for  v. 

EXAMPLE 

2^-3^  +  ^  =  4,  CO 


lx2-2x2/  +  3r  =  9.  (0) 

Putting  y=vx  in  (1)  and  (2),   2  x2  —  3  UX2  +  V2X2  =  4,  (5) 

and,  x2  -  2  VX2  +  3  t?2X2  =  9.  (4) 

Factoring  (3)  and  (4),  X2  (2  -  3  V  +  t'2)  =  4.  (5) 

x2(l  -2t7  +  3u2)=9.  (6) 


SIMULTANEOUS  QUADRATIC   EQUATIONS 

137 

Equating  the  values  of  X2  in                             4              _               9 

(5)  and  (6),                                    2  —  3  V  +  V*      1  -  2  v  +  3  v2 

•  (?) 

Clearing  (7)  of  fractions,               3  V2  +  19  V  —  14  =  0. 

(*) 

Solving  (S),                                                                  V  -—  7,  §. 

(5) 

Since  y  =  vx,                                                                  V  —  ~  '  X, 

(20) 

2x 
and,                                                                              2/  —  -^-" 

U-0 

From  U),               2  x2  +  21  x2  +  49  x2  =  72  x2  =  4. 

(12) 

'.Solving  (12),                                                             *  =  ±  T  ^2- 

6 

(13) 

From  (10),                                                                   2/  =  =F  7  ^2- 

6 

(H) 

Similarly,  from  (11)  and  (1),                                     X  =  ±  3, 

(15) 

From  (15)  and  (1/),                                                   #  =  ±  2. 

(16) 

Test  as  usual. 

„   .                                 WRITTEN    EXERCISES 
Solve : 

1.    a-2 +  ?/2  =  41,                                      9.   aj2-ajy  =  54, 
a*/  =  20.                                                 a*/  -  y2  =  18. 

2.  x2  +  xy  =  a?,  10-   ^  +  ^  =  12, 
tf  +  xy  =  b2.  tf  +  xy  =  24;. 

o      2  ,         ,     *     7  11-   4x2  +  3?/2  =  43, 

3.  x2  +  xy  +  y  =  7,  J 

4.  z2  +  ^-2</2  =  4,  12.   E±*  +  *=2'=|, 

„    n  x—y    x+y    4 

x2-3xv  +  2  =  0.  2  i     2      on 

?   '  ar  +  ?r  =  20. 

5.  x2 +6ajy +  2^  =  133,  13    o x2  +  3  xy +  f-  =  80, 

x2-y2  =  lG.  Xy-x2=6. 

6.  jb2  - 12  a#  + 119=0,  14    ^2_47/2  =  20, 
2^-2^  +  24  =  0.  ^  =  12. 

7.  a2aj2  +  6Y  =  ca,  15.   a:2  -  ay  +  ?/2  =  21, 
x*-tf  =  c?d?.  2xy-y2  =  15. 

8.  ar  +  ?/2  =  13,  16.   4  ^  +  3/ =  43, 
xy  =  6.  3ar-2/2-3  =  0. 

10 


138  ELEMENTARY   ALGEBRA 

171.    Class  IV. 

A  system  in  which  each  equation  is  unaltered  when  x  and  y  are 
interchanged  can  be  solved  by  letting  x  =  u  +  v  and  y  =  u  —  v. 

EXAMPLE 

Solve:                              l*+t-.-f-1»,  (1) 

{         xy  +  x  +  y  =  S9.  («) 

Let  x  =  u+  v,  y=u  -  v,  then  (1)  becomes 

(u  +  v)2+(u-v)2-(u  +  v)-(u-v)  =  78,  (S) 

and  (2)  becomes 

(u  +  v)(u  —  v)  +  (u  +  v)  +  (u  -  v)=39.  (4) 

Simplifying  {3),                    2  U2  +  2  V2  —  2  U  =  78.  (5) 

Simplifying  (4),                            M2  —  V2  +  2  U  =  39.  (6) 

Dividing  (5)  by  2,  and  sub-                „    „      '„      _  fl  . 

trading  (6)  from  the  result,         ^  »    —  ou  —  u.  ^/; 

Expressing  u  in  terms  oft1,                                W  —  rLil—  f^") 

3  v  J 

Substituting  (8)  in  (6),                         i^!  +  ^L  -  39.  (of) 

9        3  W 

Simplifying  (9),                  4  tf*  +  3  V2  -  9-39  =  0.  (10) 

Solving  (JO),                                                      »2  =  —  ^o  (-7-0 

and,                                                               «2  =  9.  (12) 

Solving  (ii)  and  (IS),                                       V =  ±  £  V— 39.  (i5) 

Substituting  (i3)  and  UA)  in  (S),                    v  =  ±  3,  (i\f) 

Then,                                                               «  =  -¥,  or  6.  (i5) 


-13±V-39  ,_>. 

Also                                               X  =  U+V  = ^ (16) 

=  6  ±  3  =  9  or  3,  (i7) 

-  13TV-39  ,,ON 

and,                                                      y  =  u  —  V  = — (18) 

Also                                                                =6  =F3  =  3  or  9.  (19) 

Such  equations  may  often  be  solved  by  some  of  the  previous  methods, 
in  which  case  it  is  usually  preferable  to  do  so. 


1. 

xy  —  (x  +  y)  —  l  =  0, 

xy  =  2. 

2. 

y     x     2 

M-4 

SB       ?/ 

3. 

^  +  2,2  =  39, 

y  — »  =  3. 

4. 

v*-xy  +  yi-r?  =  Q, 

x  —  y  +  l  =  0. 

SIMULTANEOUS   QUADRATIC   EQUATIONS         139 


0   ,  WRITTEN    EXERCISES 

Solve : 

5.  x2  —  xy  +  tf  =  12, 

x-  +  xy  +  f  =  4. 

6.  zy  =  3(z4-y)} 

a*  +  f  =  160. 

7.  ^  +  ?/2  +  ^  +  2/  =  188, 
a-?/  =  77. 

8.  x2-xy  +  y*  =  19, 

xy  =  15. 

9.  2x-  +  2f-(x-y)=9, 
tf  =  l.  =1. 

172.  The  foregoing  classes  of  simultaneous  quadratic  equa- 
tions are  applied  in  the  following  problems. 

WRITTEN    EXERCISES 

1.  Two  square  floors  arte  paved  with  stones  1  ft.  square ; 
the  length  of  the  side  of  one  floor  is  12  ft.  more  than  that  of 
the  other,  and  the  number  of  stones  in  the  two  floors  is  2120. 
Find  the  length  of  the  side  of  each  floor. 

Solution. 

Let  x  be  the  length  in  feet  of  a  side  of  the  smaller  floor  and  y  be  the 

length  of  that  of  the  other,  then 

x  =  y—12.  (i) 

and  by  the  given  conditions,  X2  +  y2  =  2120.  (~) 

Substituting  (i)  in  (2),       (y  -  12)2  +  y2  =  2120.  (8) 

Simplifying  (3),  y2  -  1 2  y  -  988  =  0 .  (4) 

Solving  (4),  ?/  =  38  and  —  26,  (5) 

Substituting  (5)  in  (l),  x  =  26  and  -  38.  (<?) 

The  negative  values  not  being  admissible,  the  squares  are  26  ft.  and  38  ft.  on 

a  side.  (7) 

Test.  262  +  382  =  2120. 


140  ELEMENTARY   ALGEBRA 

2.  The  sura  of  the  sides  of  two  squares  is  7  and  the  sum 
of  their  areas  is  25.     Find  the  side  of  each  square. 

3.  The  hypotenuse  of  a  certain  right  triangle  is  50,  and  the 
length  of  one  of  its  sides  is  f  that  of  the  other.  Find  the  sides 
of  the  triangle. 

4.  The  difference  between  the  hypotenuse  of  a  right  triangle 
and  the  other  two  sides  is  3  and  6, respectively.    Find  the  sides. 

5.  A  number  consists  of  two  digits ;  the  sum  of  their 
squares  is  41.  If  each  digit  is  multiplied  by  5,  the  sum  of 
these  products  is  equal  to  the  number.     Find  it. 

6.  The  difference  between  two  numbers  is  5 ;  their  product 
exceeds  their  sum  by  13.     Find  the  numbers. 

7.  In  going  120  yd.  the  front  wheel  of  a  wagon  makes  6 
revolutions  more  than  the  rear  wheel ;  but  if  the  circumference 
of  each  wheel  were  increased  3  ft.,  the  front  wheel  would  make 
only  4  revolutions  more  than  the  rear  wheel  in  going  the  same 
distance.     Find  the  circumference  of  each  wheel. 

8.  The  diagonal  of  a  rectangle  is  13  in. ;  the  difference  be- 
tween its  sides  is  7  in.     Find  the  sides. 

9.  The  diagonal  of  a  rectangle  is  29  yd.,  and  the  sum  of  its 
sides  is  41  yd.     Find  the  sides. 

10.  The  sum  of  the  perimeters  of  two  squares  is  104  ft. ;  the 
sum  of  their  areas  is  346  sq.  ft.     Find  their  sides. 

11.  The  difference  between  the  areas  of  two  squares  is  231 
sq.  in. ;  the  difference  between  their  perimeters  is  28  in.  Find 
their  sides. 

12.  Two  steamers  set  out  simultaneously  from  San  Fran- 
cisco and  Honolulu,  2100  mi.  apart,  and  travel  toward  each 
other ;  they  meet  840  mi.  from  Honolulu  in  1\  da.  less  than 
the  difference  between  their  rates  in  miles  per  hour.  Find  the 
rate  of  each  per  hour. 

13.  Two  trains  leave  New  York  simultaneously  for  St. 
Louis,  which  is  1170  mi.  distant ;  the  one  goes  10  mi.  per  hour 
faster  than  the  other  and  arrives  9|  hr.  sooner.  Find  the 
rate  of  each  train. 


SIMULTANEOUS   HIGHER   EQUATIONS  141 

SIMULTANEOUS   HIGHER   EQUATIONS 

173.    Certain  systems  containing  higher  equations  can  be 
solved  by  the  methods  of  this  chapter. 

EXAMPLES 


1.    Solve: 

f  ar5  +  y3  =18  xy, 
\x  +y  =12. 

GO 

{2) 

Put  a;  =  u  +  v,  and 

y  =  11  —  w, 

Then,  from  {1) 

(?<  +  v)3  +  (u  —  v)3=-18(?<  +  fl)(M  —  v), 

(3) 

and  from  (2), 

(u  +  v)  +  (u  -v)  =  2u  =  12. 

(4) 

Combining,    (U) 
and  (3), 

216+  18w2  =  9(36-«2). 

(-5) 

Solving  (5), 

»  =  ±2. 

(C) 

•'• 

a;  =  u  +  v  =  8  and  4, 

(7) 

and 

y  =  tt  —  v  —  4  and  8. 

(-5') 

Test  as  usual. 

2.    Solve: 

(x*  +  y4  =  706, 
U-y  =2. 

00 

09) 

Put     a?  =  u+t>, 

y  =u  — V, 
Then  from  (1) 

(m+v)4+(m-»)4=706, 

(3) 

and  from  (2), 

(?«  +  »)  —  («  —  v)=  2. 

00 

Simplifying  (&), 

0=1. 

(5) 

From  (5)  and  (3), 

(«  +  l)4  +  (?<-l)4  =  706. 

(6) 

Simplifying  (6), 

M4  +  6  ifi  -  352  =  0. 

(7) 

Solving  (7). 

u2  =  -  22. 

(*) 

and 

u2  =  16. 

(5) 

• 
•  • 

rt=  -fcV-22,  ±4. 

(20) 

x  =  u+v=±V— 22+1.    (22) 
k  =  ±  4  +  1  =  5,  -  3.      (2£) 


and,  y  =  u— f»=  ±  V-22— 1.     (25) 

t/=±4-l  =  -5,  +3.       {14) 


Test  as  usual. 


142  ELEMENTARY   ALGEBRA 

g  ,       .  WRITTEN    EXERCISES 

1.  x3  +  y3  =  lS9,  4.   x4  +  y*  =  81, 
x  +  y  =  9.  x  +  y  =  5. 

2.  x3  +  y*  =  72,  5.    x*+y*  =  2, 
x  +  y  =  6.  x  +  y  =  2. 

3.  x*  +  y*  =  l&9,  6.    x4  +  y4  =  10,001, 
tfy  +  xy1  =  180.  x  —  y  =  9. 

SUMMARY 

1.  Certain  classes  of  simultaneous  quadratic  equations  can 
be  solved  by  quadratic  methods :  Sec.  167. 

(1)  Class  I.  A  system  of  equations  such  that  substituting 
from  one  equation  into  the  other  produces  an  equation  of 
quadratic  form.  Sec.  168. 

(2)  Class  II.  A  system  in  which  one  equation  has  only 
terms  of  the  second  degree  in  x  and  y  can  be  solved  by  finding 
x  in  terms  of  y,  or  vice  versa,  from  this  equation  and  substitut- 
ing in  the  other.  Sec.  169. 

(3)  Class  III.  A  system  of  two  simultaneous  quadratic 
equations  whose  terms  are  of  the  second  degree  in  x  and  y, 
with  the  exception  of  the  absolute  terms,  can  be  solved  by 
reducing  the  system  to  one  of  Class  II. 

This  can  be  done  in  two  ways : 

Make  their  absolute  terms  alike  and  subtract.  The  result- 
ing equation  has  every  term  of  the  second  degree  in  x  and  y. 

Substitute  vx  for  y  throughout  the  equations  and  solve 
for  v.  Sec.  170. 

(4)  Class  IV.  A  system  in  which  each  equation  is  unaltered 
when  x  and  y  are  interchanged,  can  be  solved  by  letting 
x  =  u  +  v  and  y  =  u—  v.  Sec.  171. 

2.  Certain  simultaneous  higher  equations  can  be  solved  by 
quadratic  methods.  Sec.  173. 


QUADRATIC  AND    HIGHER   EQUATIONS           143 

REVIEW 

_   ,  WRITTEN    EXERCISES 

Solve : 

1.  x  +  y  =  7.5,  13.    af5  —  y3  =  665, 
xy  =  14.  x  —  y  =  5. 

2.  3x-2y  =  0,  14.    ^  +  0^  +  ^=19, 
xy  =  13.5.  a;?/ =  6. 

3.  x  +  y  =  7,  15.    (a  +  2)(z-3)  =  0, 
arJ-?/2  =  21.  x2  +  3xy  +  y2  =  5. 

4.  a;  —  ?/ =  5,  16.    a;2  —  / =  3, 

x2  +  y2  =  37.  x2  +  2/2-^  =  3. 

5.  a; -y  =1,  17     xj\-y  +  x-ji  =  5 
3x2  +  y2  =  31.  x-y      x  +  y     2' 


a?+V  =  90. 

tf  +  xy  +  y2-- 

x*  +  x*tf+yi  =  21. 

x2  +  y2-l  =  2xy, 
xy(xy  + 1)  =  8190. 


6.  x—y  =5, 

x2  +  2xy  +  y2  =  75.  18.    a^  +  agf  +  y*  :  =  7, 

7.  x  +  y  =  7(x-y), 
x2  +  y2  =  225.  19-    ar!  +  ?/2--l=2a;y, 

8.  5(x2-y2)  =  4(x2+y2), 
x  +  y  =  8. 

9.  x  +  y  =  9, 

ary         10 

Vzy      vo 

10.  ^L  =  4, 
aj*  =  20. 

11.  x  +  y  =  a, 
x3  +  ys  =  b3. 

12.  x  —  y  =  a, 
tf-f  =  b\  &  +  tf  =  5(x  +  y)+2< 


20. 

a  +  #  =  5, 

1+1  =  5. 

x     y     6 

21. 

A/                9                2 

x  +  y  =  -  =  xr  —  yr 
2/ 

22. 

»2-y2=144, 

x  —y  —  8. 

23. 

x2  +  z#  =  j^, 

•*#  +  f  =  T8  • 

24. 

144  ELEMENTARY   ALGEBRA 

25.  a?  +  tf  +  2(x  +  y)  =  12,  28.    x2  +  2 xy  +  7 /  =  24, 
xy-(x±y)  =  2.  2x2-xy-y2  =  8. 

26.  a?  +  a#  +  y2  =  21,  29.    (2a>-3)(3#-2)=0, 

a  —  V^  +  ^  =  3.  4:x2  +  12xy  —  3y2  =  0. 

Suggestion.  In  Ex.  26  divide  (1)  Suggestion.  In  Ex.  29,  from  equa- 
by  (2),  obtaining  x  +  y  +  y/xy  =  7.  tion  (1),  x  =  f.  The  correspond- 
Add  this  equation  to  (2)  and  find  ing  value  of  y  is  found  by 
x  in  terms  of  y.  substituting  this  value    of    a;   in 

equation  (2). 

27.  .r2  +  to  +  £*  =  133,  30.   x(x-y)  =  0, 

t  +  x  —  -yjtx  =  7.  x2  +  2xy  +  y2  =  9. 


31.  x—y  —  -\/x  —  y  =  2, 

a? -3/*  =  2044. 

Soggestion.  In  the  first  equation  put  Vx  —  y  =  3,  and  solve  for  z, 
finding 

x  —  y  =  1,  or  4. 

Then  from  the  second  equation, 

x2  +  xy  +  y2  =  2044,  or  511. 

Put  y  —  x  —  1  in         x2  +  xy  +  y2  =  2044,  and  solve  for  x. 

Similarly  in  the  case  of  x2  +  xy  +  y2  =  511. 

32.  3x2-4y2=8, 
5(aj-&)-4t/  =  0. 

For  what  values  of  k  are  the  solutions  real  ?     Imaginary  ? 

Equal  ? 

33.  Vic  —  Vy  =  2, 

(■y/x  —  -y/y)  'Vxy  =  30. 

34.  Two  men,  A  and  B,  dig  a  trench  in  20  days.  It  would 
take  A  alone  9  days  longer  to  dig  it  than  it  would  B.  How 
long  would  it  take  A  and  B  each  working  alone  ? 

35.  A  man  spends  $539  for  sheep.  He  keeps  14  of  the 
flock  that  he  buys,  and  sells  the  remainder  at  an  advance  of  $2 
per  head,  gaining  $28  by  the  transaction.  How  many  sheep 
did  he  buy  and  what  was  the  cost  of  each  ? 


QUADRATIC   AND   HIGHER   EQUATIONS 


145 


SUPPLEMENTARY  WORK 

Graphs  of  Simultaneous  Quadratic  Equations 
Preparatory. 
1.    Iu  the  same  diagram  construct  graphs  to  represent  the 


equations : 


x-2  +  f  =  25, 
x  —  y=—  L 


Compare  the  result  with  this  figure. 


IMI>> ]i  luAauluia) 

* "='1 

■" 

( 1 1 1 

i 

5 

/ 

;        |       | 

.. 

— ■ | :     / 

j 

/                \ 

1 

| 

\ 

— ■' 

I 

0 

-  \                ■  ./ 

: 

' ! :•• 

i     j 
i     i 

• 

L/j     1 

: 

/ 

j 

LX; 

: 

...... 

\       \       1 

2.    Solve  the  system  of  equations  in  Exercise  1. 
Compare  the  values  of  x  and  y  with  the  coordinates  of  the 
intersections  of  the  graphs. 

In  how  many  points  does  the  straight  line  intersect  a  circle  ? 
How  many  solutions  has  the  given  system  of  equations  ? 


146 


ELEMENTARY   ALGEBRA 


3.    Construct  in  one  diagram  the  graphs  of  the  aquation 

\xy=l2. 
Compare  the  result  with  this  figure. 


.1 


I      I     I'     If     f     I 


- f [■■■ 

•"X"]   ] :  T" 

1  J 

1        i 

\ r™ 

:  \     ' 
-  \    ? 

:               ;               •< 

n\         3         3 

I              -              - 

\/ 

]- 



1         1         '■ 

-/ 

^  V**^^  " 

I  [  r  i  r 

^^p>*' 

\ i  / 

1        »        ! 

o 

l               :               i 

'' \ 

F 

;•- 

>d [ 



i |" 

XX  \ 

■■••; j \ 

X 


\     \ 


4.    Solve  the  system  of  equations  in  Exercise  3. 
Compare  the  values  of  x  and  y  with  the  coordinates  of  the 
intersections  of  the  graphs. 

Graphical  Solution  of  Two  Simultaneous  Equations.  Every 
point  of  the  graph  of  one  of  the  equations  has  coordinates,  x 
and  y,  that  satisfy  that  equation ;  the  points  of  intersection 
are  points  of  both  graphs  and  therefore  have  coordinates,  x  and 
y,  that  satisfy  both  equations.  Hence,  to  solve  two  simul- 
taneous equations  graphically,  draw  their  graphs  and  read  the 
coordinates  of  their  intersections.  The  coordinates  of  each 
point  of  intersection  correspond  to  a  solution.  If  the  graphs 
do  not  intersect,  the  system  of  equations  has  no  real  roots. 


i^R 


QUADRATIC   AND   HIGHER   EQUATIONS  147 

WRITTEN    EXERCISES 
Solve  graphically,  and  test  by  computing  x  and  y. 

1.  2a2  +  y=l,  7.   xi  +  y2  =  25, 
x-y  =  2.  aj2-2/2  =  7. 

2.  a-2-?/2  =  25,  8.    x2-3xy  +  2y2  =  0, 
x  +  y  =  l.  ^  +  ^=16. 


3.  x2  +  y2  =  10,  9.   4z2  +  9?/2  =  36, 
a;y  =  3.  x2  +  y2  =  25. 

4.  a-2  +  2/2=13,  10.   4ar°  +  9/  =  36, 
xy  =  6.  2x-3y  =  5. 

5.  a^  +  r  =  13,  11.    4z2-92/2  =  30, 
a?  +  2y  =  l.  a2 +  ^  =  16. 

6.  v?-tf  =  5,  12.   4ar9-9/  =  36, 
3  a?  -  y  =  7.  a?y  =  18. 


CHAPTER   IX 

PROPORTION 

174.  Proportion.  An  equation  between  two  ratios  is  called 
a  proportion. 

Thus,  -  =  -  is  a  proportion. 

b     d 

The  older  form  of  writing  this  proportion  is  a  :  b  ::  c:d ;  whence  b 
and  c  are  called  the  means  and  a  and  d  the  extremes.  When  so  written, 
the  proportion  is  read  "  a  is  to  b  as  c  is  to  d."     At  present,  it  is  more 

customary  to  use  the  form  -  =  -  and  to  read  it  "a  over  b  equals  c  over  d." 

b      d 

ft  O 

175.  Fourth  Proportional.  In  the  proportion  -  =  -,  the 
fourth  number,  cl,  is  called  the  fourth  proportional. 

176.  Third  Proportional.     In  the  proportion  -  =  -,  the  third 

b      c 
number,  c,  although  in   the   fourth  place,  is  called  the  third 
proportional  to  a  and  b. 

177.  Mean  Proportional.  In  the  proportion  -  =  -,  b  is  called 
the  mean  proportional  between  a  and  c. 

178.  Relation  to  the  Equation.  A  proportion  is  an  equation 
and  is  to  be  treated  in  accordance  with  the  properties  of 
equations. 

EXAMPLES 

1.   Find  the  fourth  proportional  to  the  numbers  6,  8,  30. 

Let  '•  lie  the  fourth  proportional,  then,  hy  Sec.  175,  -  =  — .  (i) 

8       x 

Multiplying  by  S  x,  6  x  =  8  •  30.  (2) 

Dividing  by  fi,  £  =  40.  (3) 

Therefore  40  is  the  fourth  proportional  to  6,  8,  30. 

148 


PROPORTION  149 

2.    Find  the  third  proportional  to  5  and  17. 

Let  x  be  the  third  proportional,  then,  by  Sec.  176,  —  =  — •  (7) 

Multiplying  both  members  by  17  x,  5  X  =  172.  (,£) 

Solving  (2),  5C  =  if2  =  57|.  (5) 

Therefore  57f  is  the  third  proportional  to  5  and  17. 

WRITTEN   EXERCISES 

1.  Write  m,  n,  and  £>  so  that  p  shall  be  the  third  propor- 
tional to  m  and  n. 

2.  Write  m,  w,  and  p  so  that  m  shall  be  the  mean  propor- 
tional between  m  and  p. 

Find  x  in  each  of  the  following  proportions : 


3. 

_5_1 

x     4 

7. 

1.21       x 
x        .09 

4. 

;    IB           7 
11     1331 

8. 

1_42 
6      x 

5. 
6. 

x--16  =  i 
-  24  -=-  x  = 

:2 

a;  -T- 

-12. 

9. 

75       -a; 
—  x       3 

10.  Find  the  third  proportional  to  8  and  V— 5. 

11.  Find  the  third  proportional  to  —  6  and  V3. 

12.  Find  the  mean  proportionals  between  the  following 
numbers  :  9  and  16  ;  —  25  and  —  4 ;  V— 3  and.  V—  7 ;  1 
and  —1. 

13.  If  a  sum  of  money  earns  S48  interest  in  5  yr.,  how 
much  will  it  earn  in  16  yr.  at  the  same  rate  per  cent  ? 

14.  A  city  whose  population  was  40,000  had  2500  school 
children ;  the  total  population  increased  to  48,000,  and  the 
number  of  children  of  school  age  increased  proportionally. 
How  many  children  of  school  age  were  there  then  ? 

15.  What  number  must  be  added  to  each  of  the  four  num- 
bers, 5,  29,  10,  44,  to  make  the  results  proportional  ? 


150 


ELEMENTARY   ALGEBRA 


16.  A   lever   need   not  be  straight,  although   it 
must  be  rigid.     Thus,  the  crank  and  the  wheel  and    * 
axle  are  varieties  of  the  lever,  and  the  law  of  the 
lever  applies  to  them.     Thus,  in  Fig.  1, 

W     p' 
Find  W  if  P  =  14,  p  =  16,  and  w  =  4. 

17.  Find  the  unknown  number: 


Fig.  1. 


(1) 

(2) 

(3) 

p= 

3a 

__ 

a  —  b 

w= 

66 

5p 

(a  +  by 

p  = 

2c 

8p 

a  +  b 

w  = 

— 

2p 

— 

18.  If  an  axle  is  6  in.  in  diameter,  what  must  be  the 
diameter  of  the  wheel  in  order  that  a  boy  exerting  a  force  of 
50  lb.  may  be  able  to  raise  800  lb.  weight  ? 

19.  A  brakeman  pulls  with  a  force  of  150  lb. 
on  a  brake  wheel  16  in.  in  diameter.  The  force 
is  communicated  to  the  brake  by  means  of  an 
axle,  A,  4  in.  in  diameter.  What  is  the  pull  on 
the  brake  chain  ? 

20.  Fig.  3  represents  a  weight  W  acting  at  P  on  an 
inclined  plane,  whose  rate  of  slope  is  a  vertical  units  to  b 
horizontal  units.  It  is  known  that  the  weight  W  acts  in 
two  ways  :  a  force  N  pressing  directly  against  the  surface 
and  tending  to  produce  friction,  and  a  force  F  parallel  to 
the  plane  and  tending  to  cause  the  ^ 

weight  to  slide  down  the  plane.  It  is  known 
that  these  various  quantities  are  related  to 
each  other  thus  :  ^--' 


Z- 
w 


a 

i 


W 


Find  F  and  ^V,  if  TP=  15  lb.,  a  =  20  in.,  b  =  21  in.,  I  =  29  in. 


PROPORTION 


151 


21.  Find  I  and  N,  if  F=W  lb.,  TT=34  lb.,  a  =  4  ft.,  6  = 
1\  ft. 

22.  Find  6  and  I,  if  F=66  lb.,  ^=112  lb.,   17=130  lb., 
a  =  33  in. 

23.  Find  IF  and  a,  if  i^=200  lb.,  N=45  lb.,  &  =  9,  Z  =  41. 

24.  It  is  known  that  the  pressure  P 
exerted  by  a  letter  press  on  a  book  is 
connected  with  the  force  F  and  the  other 
dimensions  indicated  in  Fig.  1  by  the  fol- 
lowing relation : 


F_ 


h 

2  trr 


Fig.  1. 


Using  ^2-  as  it,  what  pressure 
is  exerted  by  a  force  of  20  lb.  ap- 
plied to  the  wheel  of  a  letter  press 
16  in.  in  diameter  and  the  threads 
of  whose  screw  are  ^  in.  apart  ? 

25.  It  is  known  that  if  a  circular  arch  of  span  s  is  raised  h 
units  in  the  middle,  as  indicated  in  Fig.  2,  then  h,  s,  r  satisfy 
the  proportion : 


h: 


s 
2     2 


:2r-h. 


If  a  bridge  in  the  form  of  a  circular 
arch  is  to  have  a  span  of  200  ft.  and  be 
raised  in  the  middle  6  ft.  above  the  level 
of  the  end  points,  what  is  the  radius  of 
the  circle  ? 

26.   Find  the  unknowns  : 


--%- 


V 

Fig.  2. 


(1) 

(2) 

(3) 

(4) 

h  = 

s  = 

r  = 

10 
60 

rr 

i 

287 

176 

187 

1470 
18015 

1.52 


ELEMENTARY   ALGEBRA 


27.  When  a  force  P  is  applied 
at  H,  the  dimensions  indicated  in 
the  figure  and  the  weight  W  are 
known  to  be  related  as  follows  : 


P_m 

W~ 


dr 
'2irlR 


If  72=18  in.,  r=10  in.,  1=12  in., 
d  =  \  in.,  and  -2T2  be  used  for  ■*, 
what    weight   can    a    man    raise 

who  turns  the  handle  H  with  a  force  of  90  lb.  ? 

28.    In  the  formula  of  Exercise  27  find  P,  if  W=  1000  lb., 

B  =  21  in.,  I  =  15  in.,  r  =  11  in.,  d  =  §  in. 

179.  7?t  am/  proportion  the  product  of  the  means  equals  the 
product  of  the  extremes. 

For,    -  =  - ,  and  multiplying  both  members  by  bd,   ad  =  bc. 
b     d 

180.  Conversely, 

If  the  product  of  two  numbers  equals  the  product  of  two  other 
numbers,  the  four  numbers  can  be  arranged  in  a  proportion,  the 
two  factors  of  one  product  being  the  means,  and  the  two  factors 
of  the  other  product  the  extremes. 

For,  if  ad 

b      d 


be,  divide  both  members  by  bd,  and  -  =  -■ 

b     d 


181.  If  -  =  c,then 

b      d  a      c 

For,  if  two  numbers  are  equal,  we  know  that  their  reciprocals  are 
equal.     See  First  Cocrse,  Sec.  193,  p.  141. 

The  older  form  of  statement  that  is  still  sometimes  used  is:  If  four 
numbers  form  a  proportion,  they  are  in  proportion  by  inversion. 

182.  If  -  =  -,  then  -  =  -• 

b      d  c      d 


For,   multiplying  both  members  by 


canceling  b  and  c, 


6 

c 

a  _b 
b~  c  ' 
a  _b 
c      d 

c 

or, 


The  older  form  of  statement  is :  If  four  numbers  form  a  proportion, 
they  are  in  proportion  by  alternation. 


PROPORTION  153 

183.  If  ^,then^±-6=^. 

b     d  b  d 

For,  adding  ±  1  to  both  members,  -  ±  1  =  -  ±  1. 

b  <I 

Therefore,  performing  the  processes,  a  ^  -  =  c  ^    • 

6  a" 

The  older  form  of  statement  is :  If  four  numbers  form  a  proportion, 
they  are  in  proportion  by  composition  or  by  division. 

184.  If  ^  =  ^ ,  then  «±&  =  ^±I?. 

b      d  a  —  b      c  —  d 

For,  from  Sec.  183,  £±-$  =  £+-£. 

ft  d 

a- 6     c -  «" 


and 


d 


Dividing  these  equations,  member  for  member, 

a + b _c  + d 
a  —  b     c  —  d 

The  older  form  of  statement  for  this  is :  If  four  numbers  form  a  pro- 
portion, they  are  in  proportion  by  composition  and  division. 

185.  The  mean  proportional   between  two   numbers   is   the 
square  root  of  their  product. 

For,  from  -  =  -,  Sec.  403,  we  find  b  =  Vac. 
b      c 

186.  The  properties  of  proportion  are  ^^seful  in  solving  cer- 
tain problems. 


00 

(*) 

(3) 


EXAMPLES 

olve:        Va 

+  bx  ■ 
+  bx- 

■f  Va  —  bx      1 

Va 

—  Va  —  bx     % 

By  Sec.  1S4, 

2Va  +  bx       3 

2Va  —  bx     -  l 

Simplifying  (2), 
11 

—  Va  +  bx  =  SVa  —  bx. 

154  ELEMENTARY   ALGEBRA 

Squaring  (3),  Ct  +  bx  =9<Z  —  9  &2C.  (4) 

Solving  (U),  10bx  =  8  a,  (5) 

and  CC  =  — -  •  (6) 

56 

~ .         a     c      i        4.1    j.  « 4- &      Va2 4- b2  , yx 

2.    Given  -  =  -,   show  that  — !—  =  -  •  (J) 

6     d  c  +  cZ      Vc24-d2 

By  Sec.  183,  ■    7"      =  — ^ (*) 

o  a 

.-.  by  Sec.  182,  —  =  -  •  (5) 

c  +  rf      a 

Squaring  both  members  of  the  ?L  =  £_.  (/) 

given  equation,  ^-2       (p 


Applying  Sec.  183  to  (4), 


g2  +  &2  _  C2  +  ^2 
&2  d2 


a2  +  &2  _  &2 


From  (6)i 


yqa  +  &a  _  & 
Vc2  +  d2     d 


c  +  d     v  c2  +  <72 


WRITTEN    EXERCISES 

a     &  \       fwa2  +  6'2     »2-&2 
1.    Given  -  =  -,  show  that  -        -  =  — 

be  a+  c        a  —  c 


/-i •         a      &     i,       4.1,  4.  a2  —  &2      b2—  c 
2.   Given  -  =  -,  show  that 


■> 


be  a 


3.    Given  |  =  c-,  show  that  ^^-2  =  ^^  • 


(S) 


Applying  Sec.  182  to  (5),  g2  +  ^  =  J2 '  (6) 


(7) 


Equating  values  of  £  in  (3)  and       «±& -  ^'±j! .  (5) 


PROPORTION  155 

Solve  the  equations: 

4.    V#  +  y/b  :  V#  —  s/1)  =  a  :  b.  5.    -—  =  f-  =  — - — • 

27      9      x  —  y 

6.  A  passenger  on  an  express  train  observes  that  a  train 
going  in  the  opposite  direction  passes  him  in  2  sec.  If  it 
had  been  going  the  same  way,  it  would  have  passed  in  30  sec- 
onds.    What  is  the  ratio  of  the  rates  of  the.  two  trains  ? 

187.  The  following  theorem  concerning  ratios  is  sometimes 
used : 

In  a  series  of  equal  ratios  (a  continued  proportion),  the  sum 
of  the  numerators  divided  by  the  sum  of  the  denominators  is 
equal  to  any  one  of  the  ratios. 

Proof  :  Let  -  =  -  =  —  =  ...  be  the  equal  ratios  and  r  be  their  common 
value.  b      d     f  (i) 

Then  -  =  r,-=  r,  -  =  r,  and  so  on.  (2) 

b        d         f  w 

/.  a  =  br,  c  =  dr,  e  =  fr,  and  so  on.  (5) 

.-.  adding  the  equations  in  (<?),  (a  +  c  +  e  •••)  =  (&  -j-  d  +  /--)r-  (4) 

.-.  solving  (4)  for  r,  a  +  c  +  e"'  =  r.  r5\ 

v  '  b+d+f-  ^  ' 

a  +  c  +  p  ■■•        the  sum  of  the  numerators        a       c         , 

.-.  —^ — — = ■  =  -,  or  -,  and  so  on,  since 

b  +  d+f'-'      the  sum  of  the  denominators      b       d 

r  is  any  of  these  ratios.  (6) 

188.  This  theorem  may  be  applied  to  certain  equations. 

EXAMPLES 
Solve : 


Applying  the  theorem, 

Simplifying  (2), 

Solving  (3), 
and 

Test  by  substitution. 


2  x  -  2      2 

CO 

4-9z     9a; 

2x_2_. 
4  ~9x 

(.*) 

x_  2 

2  ~  9  x 

(3) 

9x2=4, 

(4) 

x=±l 

156 


ELEMENTARY   ALGEBRA 


WRITTEN    EXERCISES 


2 


1.  Given  -  =  -  =  —,  show  that  — !— —  —  =  -■ 

b     d     f  b+d-f     d 

„     ~, .        a      c       e      i         , ,    .  a2  +  c2  +  e2      e 

2.  Given- ^-  =  ?  show  that  y  +  (P+/i  =  ? 

Va2  +  c2      a 

3.  Also  show  that  -  =  r  • 

V&  +  d      ° 

4.  Given—  ==£  =  -,  show  that-—         ^  ;         =  — . 

n       q      s  2n  —  Sq  -f-  5  s       n 

5.  Given  x  =  b-=°,  show  that  »  =  fi^±A 

a     c      d  c  +  d 

n    v   a      ■    x  —  7     5x  +  l 

6.  Find  x  in = • 

x  +  2      3x  —  2 

SUMMARY 
I.    Definitions. 

1.  An  equation  between  two  ratios  is  called  a  proportion. 

2.  The  numbers  b  and  c  are  called  the  means,  and  a  and 
d  the  extremes.  Sec.  174. 

3.  In  the  proportion  -  =  -,  the  fourth  number  d  is  called  the 
fourth  proportioned.  Sec.  175. 

4.  In  the  proportion  -  =  -,  the  third  number  c  is  called  the 
third  proportional.  Sec.  176. 

5.  In  the  proportion  -  =  -,  b  is  called  the  mean  proportional 

b      c 

between  a  and  c,  and  is  the  square  root  of  their  product. 

Sees.  177  and  185. 

II.    Properties. 

1.  A  proportion  is  an  equation  and  therefore  subject  to  the 
laws  of  equations.  Sec.  178. 

2.  In  any  proportion  the  product  of  the  means  equals  the 
product  of  the  extremes.  Sec.  17(J. 


PROPORTION  157 

3.  If  the  product  of  two  numbers  equals  the  product  of 
two  other  numbers,  the  four  numbers  can  be  arranged  in  a 
proportion,  the  two  factors  of  one  product  being  the  means, 
and  those  of  the  other  product  the  extremes.  Sec.  180. 

4.  If  -  =  c,  then  -  =  -  (inversion).  Sec.  181. 

b      d  a     c  v 

5.  If  -  =  -,  then  -  =  -  (alternation).  Sec.  182. 

b      d  c      d 

6.  If  -  =  -,  then  c±±l  =  <LAd  (composition  or  division). 

6      d  h  d  Sec.  183. 

7.  If  -  =  -,  then  ^_±A=^±_!?  (composition  and  division). 

6     d  a~b     c~d  Sec.  184. 

8.  In  a  series  of  equal  ratios  the  sum  of  the  numerators 
divided  by  the  sum  of  the  denominators  equals  any  one  of  the 
ratios.  Sec.  187. 

REVIEW 

ORAL   EXERCISES 

3     5 

1.  What  is  the  fourth  proportional  in  -  =  -? 

3     5 

2.  What  is  the  third  proportional  in  -  =  -? 

3.  What  is  the  mean  proportional  between  4  and  9  ? 

4.  State  the  principle  according  to  which  it  follows  from 

a     cthata  +  6-c+d 


d  b  d 


5.    What  is  the  value  of  x  in  -  =  -? 

a      x 


WRITTEN    EXERCISES 
Find  the  proportional  to  : 
1.    x,  y,  ab.  2.  p2,  q,  m2q.  3.    x2,  y2,  z2. 


158  ELEMENTARY   ALGEBRA 

Find  the  mean  proportional  between : 
4.    a\a2b2.  5.     -6,  4a262.  6.   27x>f,3y. 


(I       c 

When   :  =  -  show  that : 
o      a 


ac  _  a~ 

9                  9 

a-      a-  — 

b2 

c2      <?- 

d2 

a  +  2b_ 
2b 

_c+2d 
2d 

10. 


2a-3b      2c-3d 


8.    "  ="  ~"  .  11. 


9.     ^-r^^  =  ^r-".  12. 


4« 

3b 

i2- 

5b2 

3d 
4c2- 

5d2 

5  b2 
-b 

_Vc, 

5d2 

a 

;2-62 

c-cl      Vc2  -  d2 


13.  The  ratio  of  two  numbers  is  3  to  5  and  the  sum  of  their 
squares  is  544.     Find  the  numbers. 

14.  The  lengths  of  the  sides  of  a  triangle  are  in  the  ratio  of 
3  to  4  to  5  and  their  sum  is  120  ft.  What  is  the  length  of 
each  side  ? 


PROPORTION  159 


SUPPLEMENTARY  WORK 

Adding  the  same  positive  number  to  both  numerator  and 
denominator  of  a  ratio, 

(1)  Increases  the  absolute  value  of  the  ratio  if  the  numerator 
is  less  than  the  denominator ;  and 

(2)  Diminishes  the  absolute  value  of  the  ratio  if  the  numerator 
is  greater  than  the  denominator. 

For,  let  -   be  the  given  ratio  and  x  the  given  number. 
6 

To  determine  whether  -  or  a  +  x   is  the  greater,  subtract  the  second 

ratio  from  the  first. 

Then      -  —  a  ~^~  x  —  a^  ~*~  ax  ~  a^  ~  ^x  —  (a  ~  ^)x 
'    b      b  +  x~  6(6  + a;)  ~  b(b  +  x)  ' 

(1)  If  a  >  b,  the  last  ratio  is  a  positive  number,  therefore  the 

ft 

ratio  -  would  be  diminished  by  adding  x  to  each  of  its  terms, 
b 

which  proves  (2). 

(2)  If  a  <  b,  the  last  ratio  is  a  negative  number,  therefore 
the  ratio  -  would  be  increased  by  adding  x  to  each  of  its  terms, 

which  proves  (1). 

(3)  If  a  =  b,  the  last  ratio  is  zero,  therefore  the  ratio  -  would 
be  unchanged  by  adding  x  to  each  of  its  terms. 


ORAL   EXERCISES 

1.  How  is  the  value  of  |  changed  by  adding  4  to  each  term  ? 

2.  How  is  the  value  of  |  changed  by  adding  5  to  each  term  ? 

3.  How  is  the  value  of  4  changed  by  adding  a  to  each  term  ? 

4.  How  is  the  value  of  |  changed  by  adding  b  to  each  term  ? 

5.  What  is  the  effect  on  £  of  adding  c  to  each  term  ? 


CHAPTER   X 

VARIATION 

189.  Direct  Variation.  When  two  variable  quantities  vary 
so  as  always  to  remain  in  the  same  ratio,  each  is  said  to 
vary  directly  as  the  other.  Each  increases  or  decreases  at  the 
same  rate  that  the  other  increases  or  decreases. 

Consequently,  if  x  and  y  are  two  corresponding  values  of  the 
variables  and  k  the  fixed  ratio,  then 

-  =  k  and  x  —  ky. 
V 

For  example,  at  a  fixed  price  (k)  per  article,  the  total  cost  (x)  of  a 

number  of  articles  of  the  same  sort  varies  directly  with  the  number  of 

x 
articles  (y).    That  is,  -  =  k. 

Likewise  in  the  case  of  motion  at  a  uniform  rate  (r),  the  distance  tra- 
versed (d)  varies  as  the  time  of  motion  (t) .     That  is,  -  =  r. 

190.  A  symbol  still  occasionally  used  for  "  varies  as  "  is  oc 

Tims,  ux  varies  as  y  "  is  written  x  oc  y, 
and  "  d  varies  as  £"  is  written  dazt. 

191.  Relation  of  Variation  to  Proportion.  When  one  vari- 
able varies  directly  as  another,  any  pair  of  values  of  the  vari- 
ables forms  a  proportion  with  any  other  pair. 

For,    —  =  rand —  =  r;     .\  —  =  — ,  which  is  a  proportion. 

yl  yll  yl  yll 

192.  Expressions  for  Direct  Variation.  We  have  thus  seen 
that  the  relation  x  varies  directly  as  y  may  be  expressed  in  any 
one  of  three  ways  : 

(a)  x  =  ky,  by  use  of  the  equation. 

(b)  xcxy,  by  use  of  the  symbol  of  variation. 

(c)  —  =  — . ,  by  use  of  the  proportion  ;  x',  y'  and  x",  y"  being 

y     y 

any  two  pairs  of  corresponding  values  of  the  variables. 

160 


VARIATION  161 

WRITTEN     EXERCISES 

1.  Write  the  statement  "v  varies  as  w"  in  the  form  of  an 
equation,  also  in  the  form  of  a  proportion. 

2.  Write  the  statement  x  =  ky  by  use  of  the  symbol  oc ; 
also  in  the  form  of  a  proportion. 

t'       t" 

3.  Write  —  =  7^77  by  use  of  the  symbol  oc. 

4.  The  weight  (w)  of  a  substance  varies  as  the  volume  (v) 
when  other  conditions  are  unchanged.  Express  this  law  by  use 
of  the  equation.  By  use  of  the  symbol  oc.  Also  in  the  form  of 
a  proportion. 

5.  In  the  equation  w  =  kv,  if  to  =  4  and  v  =  2,  what  is  the 
value  of  k  ?  Using  this  value  of  k,  what  is  the  value  of  w 
when  v  =  25? 

6.  When  1728  cu.  in.  of  a  substance  weigh  1000  ounces, 
what  is  the  ratio  of  the  weight  (w)  to  the  volume  (v)  ?  What 
volume  of  this  substance  will  weigh  5250  ounces  ? 

7.  The  cost  (c)  of  a  grade  of  silk  varies  as  the  number  of 
yards  (n).     Find  the  ratio  (r)  of  c  to  n  when  c  is  $  7.00  and 


c  i        i         c' 


8.    If  in  Exercise  7,  -  =r,  what  does  —  equal  ?     Given  that 

u  n 

40  yd.  of  silk  cost  $  60,  find  the  cost  of  95  yd.  by  means  of  the 

c'      c 
proportion  —  =  - .     Also  by  means  of  the  equation  c  =  nr. 

193.  Inverse  Variation.     A  variable  x  is  said  to  vary  inversely 

as  a  variable  y,  if  it  varies  directly  as  -■ 

Inverse  variation  means  that  when  one  variable  is  doubled  the  other 
is  halved  ;  when  one  is  trebled  the  other  becomes  \  of  its  original  value, 
and  so  on. 

194.  Expressions  for  Inverse  Variation.  The  relation  "x 
varies  inversely  as  y "  may  be  expressed : 

(a)  By  the  equation,  x  =  Jc(-\  or   x  =   -,  .-.  xy  =  k. 


162  ELEMENTARY   ALGEBRA 


(b)  With  the  symbol  of  variation,  y  oc  -■ 

(c)  As  a  proportion  ~  =  ^-T. 


WRITTEN     EXERCISES 

1.  Write  the  statement  "v  varies  inversely  as  w"in  the 
form  of  an  equation.     Also  in  the  form  of  a  proportion. 

2.  Write  t  =  -  by  use  of  the  symbol  oc,  also  in  the  form  of 
a  proportion. 

3.  In  a  bicycle  pump  the  volume  (v)  of  air  confined  varies 
inversely  as  the  pressure  (p)  on  the  piston.  Write  the  rela- 
tion between  v  and  p  in  three  ways. 

4.  In  Exercise  3,  if  v  =  18  (cu.  in.)  and  p  =  15  (lb.),  what 
is  k  in  v  —  -  ?    What  is  the  pressure  ( p)  when  v  =  1  (cu.  in.)  ? 

5.  In  an  auditorium  whose  volume  (c)  is  25,000  cu.  ft. 
there  are  2000  persons  (p).  What  is  the  number  (h)  of  cubic 
feet  of  air  space  to  the  person  ?  What  will  be  the  number 
when  1000  more  persons  come  in  ? 

6.  The  area  of  a  triangle  varies  as  the  base  times  the  alti- 
tude. If  the  area  is  12  when  the  base  is  8  and  the  altitude  3, 
what  is  the  area  of  a  triangle  whose  base  is  40  and  altitude  20  '.' 

7.  The  area  of  a  circle  varies  as  the  square  of  its  radius. 
The  area  of  a  circle  of  radius  2  is  12.5664;  what  is  the  area  of 
a  circle  whose  radius  is  5.5  ? 

8.  The  volume  of  a  sphere  varies  as  the  cube  of  its  radius. 
If  the  volume  of  a  sphere  whose  radius  is  3  is  113.0976,  what 
is  the  volume  of  a  sphere  whose  radius  is  5  ? 

9.  If  x  oc  y  and  x  =  6  when  y  =  2,  find  x  when  y  =  S. 

Suggestion,      x  =  ky.      .-.  6  =  k  •  2  and  k  =  3.      Substitute  y  —  8  in 
the  equation  x  =  3  y. 

10.  Determine  k  in  xccy,  if  .r  =  10  when  y  =  20.     Also  if 
x  =  1  when  y  =  5.     If  x  =  100  when  y  =  10. 

11.  If  x  oc  iv  and  y  cc  w,  prove  that  x  4-  y  x  w. 


VARIATION 


163 


12.  If  x  oc  w  and  w  oc  y,  prove  that   xy  oc  iv2. 

13.  If  x oc  ii  and  *o oc z,  prove  that  —  oc^  • 

w      z 

14.  If  xccy-z  and  a  =  l  when  ?/  =  2  and  z  =  3,  find  the  con- 
stant A;. 

15.  Given  y  =  z-\-iv,  zccx  and  miock;  and  that  sc  =  l  when 
w  =  6,  and  that  x  =  2  when  z  =  20.    Express  y  in  terms  of  x. 

Suggestion,     y  =  kx  +  k'x  ;  determine  k  and  k'. 

16.    Given  z<xx  +  y  and  yccx2,  and  that  ic=|when  y=^ 
and  z  =  \.     Express  z  in  terms  of  x. 


GRAPHICAL   WORK 

195.  The  relation  "oj  varies  as  y"  has  been  expressed  by  the 
linear  equation  x  =  ky,  and  has  been  represented  by  a  straight 
line  (see  Eirst  Coukse,  Sees.  212  and  213,  pp.  168-170). 

The  relation  "  x  varies  inversely  as  y  "  has  been  expressed  by 

k 
the  equation  x  =  -  •     The  graph  of  this  equation  is  a  curve. 

y 

2 
Suppose,  for  illustration,  that  k  =  2.    Then  x  =  -  •    For  this  equation  : 

y 


The  table  is 


X 

2/ 

5 

2 
5 

4 

1 
2 

3 

2 
3 

2 

1 

1 

2 

i 

2 

4 

-1 

— 

2 

-2 

— 

1 

-  3 

— 

2 
3 

-4 

— 

1 
2 

-5 

— 

2 

5 

The  graph  is 


y 


j- 


r j" 

i 



4- 

j      4 

« •■ 

j 

j      3 

..; 

■4 ; 

[ |" 

1      2 

1 

: 

- 

: 

-4:     -3; 

-2)    - 

ii     0 

1 

:2 

13 

|4      ; 

! [•■ 

1 -1 

...1 3 

\-2 
V3 

j h 

T'-4 

■■- 

-j | 

• 

_J_ 

'"1 

y 


164 


ELEMENTARY   ALGEBRA 


WRITTEN   EXERCISES 
Kepresent  graphically : 


1.  x  =  -,  or  xy  =  1. 

y 

_  2 

2.  a;  =  — -,  or  xy  =  —  2. 


3.  #  =  -,  or  a-?/ =  4. 

4.  3»=  ^— ,  or  3xy  =  —  1. 


2/  2/ 

5.    Under  standard  conditions  the  volume  (V)  of  a  confined 

gas  varies  inversely  as  the  pressure  (jp).     That  is,  v  =  -.    This 

is  known  as  Boyle's  Law.     Suppose  when  the  piston  is  at  a  in 

B     the  figure,  the  volume  of  the 
gas   in   part   A    is    1    cu.    ft., 


P     P 


and     the    pressure    at    P   is 

5   lb.      When  the  pressure  is 

371  10   lb.  the  piston  moves  to  b, 

A        and  the  volume  of  the  gas  B 

becomes  \  cu.  ft.     What  will  be  the  volume  of  the  gas  when 

P=20  lb.? 

k 

Taking  k  =  5  in  the  equation  v  =  -,  the  table  of  values  is 

P 


p  = 

1 

2 

3 

4 

5 

G 

7 

8 

9 

10 

15 

20 
i 

4 

V  = 

5 

2* 

l3 

n 

1 

5 
6 

5 
7 

5 
8 

5 
9 

1 

i 

3 

The  graph  of  the  table  is  : 


I     v 


i r r 


5 

J   4 

- :  - : ~.:  :-.^  >     .-  ;- ..  ■  = ; 

3 

i 

{.nil 

2 

...|.^vi. : 1 | ■ 1 1 i ! j i L ! | ! i ..j 

|_1 

\^_____L_L:! 

i     ° 

I 

i 

.2     .3     |4     |6       6     :7       8     jB       10     11     12     13   ;14   .15    15     1?     13     19     20    ' 

: 

...j ! | L                      ,.,  ;..;....! | \ | \ 

I    i    i 


Read  from  the  graph  the  pressure  on  the  piston  necessary  to  hold  this 
volume  of  gas  at  |  cu.  ft.  ;  at  |  cu.  ft. ;  at  5  cu.  ft. ;  at  2  cu.  ft. 


VARIATION 


165 


6.  The  attraction  or  "  pull  "  of  the  earth  on  bodies  in  its  neighborhood 
is  the  cause  of  their  weight.  The  law  of  gravitation  states  that  the  weight  of 
a  given  body  varies  inversely  as  the  square  of  its  distance  from  the  center 
of  the  earth.     This  law  may  be  expressed  by  the  equation 

k 
w  =  —  • 
d? 

To  construct  the  graph  which  shows  the  nature  of  the  relation  between 
to  and  d  as  they  vary,  k  may  be  taken  to  be  1. 

Fill  the  blanks  in  the  table : 


w 


±1 

±5 

±i 

±.25 

±•2 

1 

(     ) 

(     ) 

(     ) 

(     ) 

7.   Plot  the  graph  for  the  table  in  Exercise  6. 


SUMMARY 


I.    Definitions. 


1.  When    two  variable    quantities  vary  so    as   always   to 
remain  in  the  same  ratio,  each  varies  directly  as  the  other. 

Sec.  189. 

2.  When  one  variable  varies  directly  as  another,  any  pair  of 
values  of  the  variables  forms  a  proportion  with  any  other  pair. 

Sec.  191. 

3.  A  variable  x  varies  inversely  as  a  variable  y,  if  it  varies 


directly  as  -  • 


Sec.  193. 


II.    Processes. 

1.  Processes   involving  direct   variation  are   performed  by 
reference  to  the  equation  x  =  ky.  Sec.  192. 

2.  Processes  involving  inverse  variation  are  performed  by 


reference  to  the  equation  x  = 


Sec.  194. 


y 


3.    Direct  and  inverse  variation  may  be  represented  graphi- 
cally. Sec.  195. 


166 


ELEMENTARY   ALGEBRA 


SUPPLEMENTARY   WORK 

A  large  number  of  problems  in  science  may  be  solved  by  the 
following  plan : 

EXAMPLES 

1.  The  "law  of  gravitation"  states  that  the  weight  of  a  given 
body  varies  inversely  as  the  square  of  its  distance  from  the 
center  of  the  earth.  What  is  the  weight  of  a  body  5  mi.  above 
the  surface  of  the  earth,  which  weighs  10  lb.  at  the  surface 
(4000  mi.  from  the  center)  ? 

Method.  There  are  two  variables  in  the  problem,  the  weight  (w)  and 
the  distance  (d).  There  are  also  two  parts  or  cases  in  the  statement,  one 
in  which  the  value  of  one  variable  is  unknown,  and  one  in  which  the 
values  of  both  variables  are  given. 

Arrange  the  data  as  follows  : 


10 

d 

1st  case 
2d  case 

X 

10 

4005 
4000 

To  this  table  apply  the  law  of  variation  expressed  in  the  physical  law. 

Since  the  law  is  :  w  varies  inversely  as  the  square  of  d,  the  values  of  d 

must  be  squared,  and  the  ratio  x  :  10  equals  the  inverse  ratio  of  40052  to 

40002  ;  that  is, 

x  _  40002 

10_40052' 
.-.  x  =  9.96,  and  the  weight  is  9.96  lb. 

2.  The  squares  of  the  times  of  revolution  of  the  planets 
about  the  sun  vary  directly  as  the  cubes  of  their  distances 
from  the  sun.  The  earth  is  93,000,000  mi.  from  the  sun, 
and  makes  a  revolution  in  approximately  365  da.;  what  is 
the  distance  of  Venus  from  the  sun,  taking  its  time  of  revolu- 
tion to  be  226  da.  ? 


VARIATION  10" 


Solution 


t  =  time  of  re  vol. 

d  =  distance  from  sun 

1st  case 
2d  case 

365 

226 

93,000,000 

X 

According  to  the  astronomical  law  the  times  must  be  squared  and  the 
distances  cubed  ;  then,  since  the  law  is  that  of  direct  variation : 

365'2  _  93,000,0003 
2262  ~  Xs 

\2 


.-.  x  =  93,000, 000-^/  f—\  =68,900,000,  and  the  distance  of  Venus 
from  the  sun  is  approximately  69,000,000  miles. 

WRITTEN    EXERCISES 

1.  The  intensity  of  light  from  a  given  source  varies  in- 
versely as  the  square  of  the  distance  from  the  source.  If  the 
intensity  (candle  power)  of  the  light  from  an  electric  lamp 
is  4  at  a  distance  of  150  yd.,  what  is  its  intensity  at  a  dis- 
tance of  25  yd.  ? 

2.  According  to  the  first  sentence  of  Exercise  1,  how  much 
farther  from  an  electric  light  must  a  surface  be  moved  to 
receive  only  i  as  much  light  as  formerly? 

3.  The  time  of  oscillation  of  a  pendulum  varies  directly  as 
the  square  root  of  its  length.  What  is  the  length  of  a  pendu- 
lum which  makes  an  oscillation  in  5  sec,  a  2-second  pendulum 
being  156.8  in.  long? 

4.  According  to  Exercise  3,  what  is  the  time  of  oscillation 
of  a  pendulum  784  in.  long  ? 

5.  The  distance  through  which  a  body  falls  from  rest  varies 
as  the  square  of  the  time  of  falling.  A  body  falls  from  rest 
576  ft.  in  6  sec. ;  how  far  does  it  fall  in  10  sec.  ? 

6.  Volumes  of  similar  solids  vary  as  the  cubes  of  their 
linear  dimensions.  The  volume  of  a  sphere  of  radius  1  in.  is 
4.1888  cu  in.;  what  is  the  volume  of  a  sphere  whose  radius  is 
5  in.  ? 


168 


ELEMENTARY   ALGEBRA 


7.  According  to  Exercise  6,  what  is  the  radius  of  a  sphere 
whose  volume  is  33.5104  cu.  ft.  ? 

8.  According  to  Exercise  6,  if  a  flask  holds  \  pt.,  what  is 
the  capacity  of  a  flask  of  the  same  shape  4  times  as  high  ? 

9.  In  compressing  a  gas  into  a  closed  receptacle,  as  in  pump- 
ing air  into  an  automobile  tire,  the  pressure  varies  inversely 
as  the  volume.  If  the  pressure  is  25  lb.  when  the  volume  is 
125  cu.  in.,  what  is  the  pressure  when  the  volume  is  115  cu.  in.  ? 

10.  According  to  Exercise  9,  if  the  pressure  is  50  lb.  when 
the  volume  is  250  cu.  in.,  what  is  the  volume  when  the  pres- 
sure is  10  lb.  ? 

11.  It  is  known  that  if  one  gear  wheel  turns  another  as  in  the  figure, 

the  number  of  revolutions  of  the  two 
are  to  each  other  inversely  as  their 
number  of  teeth.  That  is,  if  the  first 
has  C\  teeth  and  makes  Bi  revolutions, 
and  the  second  has  C'z  teeth  and  makes 
in  the  same  time  i?2  revolutions, 


then 

Find  R,  if  d  =  25,  C,  =  15,  and  R,  =  6. 
Find  the  numbers  to  fill  the  blanks : 


R\ Co 

R-2       C\ 


(i) 

(2) 

(3) 

(4) 

Ci  = 

42 

60 

5  n 

— 

a  = 

— 

50 

3  n 

40 

#i  = 

12 

— 

21 

8n 

R>  = 

9 

15 

— 

6  n 

12.  A  diamond  worth  $  2000  was  broken  into  two  parts,  to- 
gether worth  only  $  1600.  If  the  value  of  a  diamond  is  propor- 
tional to  the  square  of  its  weight,  into  what  fractions  was  the 
original  diamond  broken  ?     (Find  result  to  nearest  hundredth.) 

13.  If  the  volume  of  a  sphere  varies  as  the  cube  of  its  radius, 
find  the  radius  of  a  sphere  -whose  volume  equals  that  of  the  sum 
of  two  spheres  whose  radii  are  respectively  6  ft.  and  3.5  ft. 


VARIATION  169 

14.  The  number  of  vibrations  (swings)  made  by  two  pen- 
dulums in  the  same  time  are  to  each  other  inversely  as  the 
square  roots  of  their  lengths.  If  a  pendulum  of  length 
39  in.  makes  1  vibration  per  second  (seconds  pendulum),  about 
how  many  vibrations  will  a  pendulum  10  in.  long  make  ?  How 
long  must  the  pendulum  be  to  make  10  vibrations  per  second  ? 

15.  Two  towns  join  in  building  a  bridge  which  both  will  use 
and  agree  to  share  its  cost  ($  5000)  in  direct  proportion  to 
their  populations  and  in  inverse  proportion  to  their  distances 
from  the  bridge.  One  town  has  a  population  of  5000  and  is 
2  mi.  from  the  bridge;  the  other  has  a  population  of  9000 
and  is  6  mi.  from  the  bridge.     What  must  each  pay  ? 

16.  If  a  :  b  =p  :  q,  prove  that 


a2    .    h2  .        «        _  „2    i_  Q2  .      P        . 

a    1   u   .              — p  -\-  (/   . 

a  +  b                   P+Cl 

Suggestion. 

The  given  proportion  may  be  written : 

a  _b 
P~ 1 

Let 

-  =  r,  then  a  =  pr,  b  =  qr. 

p 

The  proportion  to  be  proved  may  be  written  : 

(a  +  h)(a*  +  ftg)  =  (p  +g)(;>2+  92) 
a3  p3 

Substituting  the  values  of  a  and  b  above,  the  left  member  readily 
reduces  to  the  right. 

Note.  A  good  method  for  proving  such  identities  is  to  begin  with  the 
required  relation  and  transform  it  into  the  given  relation,  or  to  transform 
both  the  given  and  the  required  relation  until  they  reduce  to  the  same 
thing. 

17.   Solve,  using  the  principle  of  composition  and  division, 

(a  -  V2bx  +  x2)  :  (a  -  b)  =  (a  +  V2  bx  -f-  x2)  :  (a  +  b). 

18.       If    «1  =  «i  =  «?  =  .  .  .    =  On      prQve  that 

&i      &2      b3  bn 

q1  +  «g  +  «3-l \-aH  _  <h  § 

h+h  +&3+  "•  +  &„         &/ 

19.    Given  x  +  -y/x  :  x  —  V#  : :  3  Vcc  +  6  :  2  Vai,  to  find  a;. 
12 


CHAPTER   XI 

SERIES 

196.  Series.  If  a  set  of  numbers  is  specified  according  to 
some  law  or  system,  the  set  of  numbers  is  called  a  series. 

197.  Terms.  The  numbers  constituting  the  series  are  called 
its  terms,  and  are  named  from  the  left,  1st  term,  2d  term,  etc. 

The  following  are  examples  of  series : 

1.  1,2,3,4,5,.-..  7.  1, 1,  |,  j,  |,  .... 

2.  1,  3,  5,  7,  9,  ••..  8.  1,  3,  9,  27,  81,  243,  .... 

3.  1,5,9,13,17,.-..  9.  2,f,f,^,  .... 

4.  3,6,9,12,15,-...  10.  1,1,1,1,1,1,.... 

5.  1,  li  2,  21  3,  ....  11.  V2,  V3,  V4,  V5,  V6,  V7,  •••• 

6.  2,  4,  8,  16,  32,  ....  12.  100,  99,  98,  97,  96,  95,  .... 

ORAL    EXERCISES 
1-12.    State  the  next  five  terms  of  each  series  above. 

198.  When  the  laiv  of  a  series  is  known,  any  term  may  be 
found  directly. 

EXAMPLES 

1.  The  law  of  the  second  series  in  Sec.  197  is  that  each  term  is  two 
more  than  the  preceding.  To  get  the  tenth  term  we  start  from  1  and 
add  2  for  each  of  9  terms.     That  is,  the  tenth  term  is  1  +  9  •  2  =  19. 

Similarly,  the  12th  temi  is  1  +  11  •  2  =  23, 
the  loth  term  is  1  +  14  •  2  =  29, 
the  47th  term  is  1  +  46  .  2  =  93, 
the  nth  term  is  1  +  (n  —  1)2  =  2 n  —  1. 

2.  The  law  of  the  eighth  series  is  that  each  term  after  the  first  is  3 
times  the  preceding  term.  To  get  the  ninth  term  we  start  at  1  and  multiply 
by  3  eight  times,  or  by  3s.     That  is,  the  ninth  term  is  1  •  38  =  G561. 

Similarly,  the  12th  term  is  1  •  3"  =  177,147, 
the  nth  term  is  1  •  3"-1  =  3»-i. 

170 


SERIES  171 

WRITTEN    EXERCISES 

1-4.  Select  four  of  the  series  in  Sec.  197  that  can  be  treated 
like  the  first  example  and  write  the  10th,  12th,  15th,  and  47th 
terms  of  each. 

5-8.  Select  four  of  the  series  in  Sec.  197  that  can  be  treated 
like  the  second  example  and  write  the  8th,  10th,  and  ?ith  terms 
of  each. 

9.  Write  similarly  the  7th,  11th,  20th,  47th,  and  «th  term 
for  any  6  of  the  above  series. 

199.  We  shall  give  only  two  types  of  series,  the  arithmetical 
and  the  geometric,  the  laws  of  which  are  comparatively  simple. 

ARITHMETICAL  SERIES 

200.  Arithmetical  Series.  A  series  in  which  each  term  after 
the  first  is  formed  by  adding  a  fixed  number  to  the  preceding 
term  is  called  an  arithmetical  series  or  arithmetical  progression. 

201.  Common  Difference.  The  fixed  number  is  called  the 
common  difference,  and  may,  of  course,  be  negative. 

For  example  : 

1.  7,  15,  23,  31,  39,.  •••  is  an  arithmetical  series  having  the  common  dif- 
ference 8. 

2.  16,  14^,  13,  111,  io,  ...  is  an  arithmetical  series  having  the  common 
difference  —  f. 

ORAL    EXERCISES 

1.  Name  all  the  arithmetical  series  in  the  list,  Sec.  197. 

2.  State  the  common  difference  in  each  of  these  series. 

3.  State  the  10th  term  in  each  of  these  series. 

WRITTEN    EXERCISES 

1.  Write  the  nth  term  in  each  arithmetical  series  in  the 
list,  Sec.  197. 

2.  Beginning  with  2  find  the  100th  even  number. 

3.  Beginning  with  1  find  the  100th  odd  number. 


172  ELEMENTARY   ALGEBRA 

4.  Beginning  with  3  find  the  200th  multiple  of  3. 

5.  A  city  with  a  population  of  15,000  increased  600  persons  per 
year  for  10  yr.    What  was  the  population  at  the  end  of  10  yr.  ? 

202.  A  general  form  for  an  arithmetical  series  is : 

a,  a  +  d,  a  +  2 d,  a  +  3 d,  ■■-,  a  +(n - T)d,  •••, 
where  a  denotes  the  first  term, 

d  denotes  the  common  difference,  and 
n  denotes  the  number  of  the  term. 

203.  Last  Term.  If  the  last  term  considered  is  numbered  n 
and  denoted  by  I,  we  have  for  the  last  of  n  terms  the  formula : 
l  =  a+{n  —  l)d. 

204.  The  Sum  of  an  Arithmetical  Series.  The  sum  of  n  terms 
of  an  arithmetical  series  can  be  found  readily. 

EXAMPLE 
Find  the  sum  of  the  first  6  even  numbers. 

1.  Let  8  =  2  +  4  +  6  +  8  +  10+12. 

2.  We  may  also  write  s-  12  +  10  +  8  +  6  +  4  +  2. 

3.  Adding  (1)  and  (2), 

2s  =  (2  +  12)  +  (4  +  10)  +  (6  +  8) +(8  +  6)  +  (10  +  4) 

+  (12  +  2) 
=  6(2  +  12)  ;  for  each  parenthesis  is  the  same  as  2  +  12. 

4.  .,s  =  6(2_±i2)=42. 

WRITTEN    EXERCISES 

Find  similarly  the  sum  of : 

1.  The  first  6  odd  numbers. 

2.  The  first  6  multiples  of  3. 

3.  The  first  5  multiples  of  7. 

4.  The  first  4  multiples  of  8. 

205.    General  Formula  for  the  Sum.     The  general  form  of  the 
series  may  be  treated  in  the  same  way. 

If  I  denotes  the  last  of  n  terms,  the  term  before  it  is  denoted 


SERIES  173 

by  /  —  d,  the  next  preceding  by  1  —  2  d,  and  so  on.     Hence,  the 
sum  of  n  terms  may  be  written  : 

s=a+(a+d)+(a+2d)  +  •••  +  (/  -2  d)  +  (l  -  d)  +  l. 
And  also,  s=  I  +  (I -d)  +  (1-2 d)  +  —  +  (a+2d)  +  (a+d)+a. 

Whence,  adding,  2s=(a  +  l)  +  (a  +  l)  A \-(a+l)  =  n(a+l). 

Therefore,  s  =  w^a       '  • 

Or,  in  words, 

T%e  sum  of  any  number  of  terms  of  an  arithmetical  series  is  one, 
half  the  sum  of  the  first  and  the  last  terms  times  the  number  of 
terms. 

By  using  the  value  of  I  in  Sec.  203,  s  =  n^2a  +  fa-1)*1'). 

This  permits  the  calculation  of  s  without  working  out  separately  the 
value  of  I. 

WRITTEN    EXERCISES 

For  each  series  in  the  following  list  find  :  First,  the  sum  of 
10  terms.     Second,  the  sum  of  n  terms. 

1.  1,2,3,4,5,—.  4.   3,6,9,12,15,—. 

2.  1,3,5,7,9,-...  5.   1,  li  2,  2|,  3,.-.. 

3.  l,5j  9,13,17,.-..         6.    100,  99,  98,  97,  96,  95,- ••. 

7.  A  man  invests  $100  of  his  earnings  at  the  beginning  of 
each  year  for  10  yr.  at  6%,  simple  interest.  How  much  has 
he  at  the  end  of  10  yr.  ? 

Solution. 

1.  The  last  investment  bears  interest  1  yr.  and  amounts  to  $106  ;  the 
next  to  the  last  bears  interest  2  yr.  and  amounts  to  $112,  etc. ;  the  first 
bears  interest  10  yr.  and  amounts  to  $  160. 

2.  Hence,  a=$  106,  d  =  $6,  and  n  =  10. 

3.  Therefore,  I  =  106  +  9  •  6  =  160. 

4.  Therefore,  s  =  V°-  (106  +  160)  =  1330. 

5.  The  man  has  $1330. 

8.  If  $  50  is  invested  at  the  beginning  of  each  year  for  20  yr. 
at  5  %  simple  interest,  what  is  the  amount  at  the  end  of  20  yr.  ? 


174  ELEMENTARY   ALGEBRA 

9.  If  a  body  falls  approximately  16  ft,  the  first  second  and 
32  ft.  farther  in  each  succeeding  second,  how  far  does  it  fall 
in  5  sec.  ? 

206.  Collected  Results.  The  three  chief  formulas  of  arith- 
metical series  are  : 

1.  l  =  a+(n-l)d. 

2.  s-        2 

"     3_    s  =  n[2a  +  (n-l)<Z\. 

GEOMETRIC  SERIES 

207.  Geometric  Series.  A  series  in  which  each  term  after  the 
first  is  formed  by  multiplying  the  preceding  term  by  a  fixed 
number  is  called  a  geometric  series,  or  a  geometric  progression. 

208.  Common  Ratio.  The  fixed  multiplier  is  called  the  com- 
mon ratio,  and  may  be  negative. 

For  example : 

1.  2  is  the  common  ratio  in  the  geometric  series  2,  4,  8,  16,  •••. 

2.  — i  is  the  common  ratio  in  the  geometric  series,  27,  —9,  3,  —1,  J,  •••. 

ORAL   EXERCISES 

1.  Name  all  the  geometric  series  in  the  list,  Sec.  197. 

2.  State  the  common  ratio  in  each  of  these  series. 

3.  State  the  6th  term  of  each  of  these  series. 

WRITTEN    EXERCISES 

1.  The  series  1,  3,  9,  27,  81,  •••,  whose  ratio  is  3,  is  the  same 
as  1,  31,  32,  33,  34,  •••.  Write  by  use  of  exponents  the  6th  term 
of  this  series ;  the  8th  term  ;  the  10th  term ;  the  15th ;  the 
25th;  the  100th. 

2.  The  series  3,  -#,  f,  —  f,  ••■,  whose  ratio  is  —  %,  is  the 
same  as  3,  3(-£),  3(-£)2,  3(-i)3,  ••-.  Write  by  use  of  ex- 
ponents the  5th  term  of  this  series ;  the  8th  term  ;  the  10th ; 
the  25th  ;  the  50th. 


SERIES  175 

209.    A  general  form  for  the  geometric  series  is: 


,.n-l 

"j 


'> 


where  a  denotes  the  first  term, 

r  denotes  the  common  ratio,  and 
n  denotes  the  number  of  the  terms. 

210.  Last  Term.    If  the  last  term  is  numbered  n,  and  de- 
noted by  I,  then  we  have  for  the  last  of  n  terms  the  formula, 

l  =  arn-\ 

211.  The  Sum  of  a  Geometric  Series.    The  sum  of  n  terms  of  a 
geometric  series  can  readily  be  found. 

EXAMPLE 

Find  the  sum  of  5  terms  of  the  series  2,  6,  18,  54,  162. 

Solution. 

Let                                           s  =  2  +  6  +  18  +  54  +  162.  (/) 

■"aSS  & the                3s  =  6  +  18  +  54  +  162  +  486.  (« ) 

Subtracting  (1)  from  (2),  3  S  -  S  =  486  —  2,  (3) 

or,                                           2s  =  484.  (4) 

Dividing  by  2,                                  S  =  242.  (5) 

WRITTEN    EXERCISES 
Find  similarly  the  sum  of  5  terms  of  each  of  these  series: 

1.  6,30,150,-.  3.    i,^,^,.... 

2.  7,-14,28,.-..  4.    i   -i,  iV,  •••• 

212.  General  Formula  for  the  Sum.     The  general  form  of  the 
series  may  be  treated  in  the  same  way.     If  I  denote  the  last 

of  n  terms,  the  term  before  it  is  denoted  by  -,  the  next  preceding 

r 

by—,  and  so  on.     Hence,  the  sum  of  n  terms  may  be  written : 

1.    s  —  a  +  ar  +  «r  + --\ \-l. 

r      r 


176  ELEMENTARY   ALGEBRA 

2.  Then  rs  =  ar  +  ar2  + -\ \- I  +  Ir. 

r2     r 

3.  Subtracting,      s  —  rs  =  a  —  Ir. 

4.  Or,  (1  —  r)  s  =  a  —  Zr. 

a  —  Ir      Ir  —  a 

'  .*.s  =  - = -• 

1  —  r        ?*  —  1 

In  words, 

T7te  s?*m  o/  any  number  of  terms  of  a  geometric  series  is  the 
ratio  times  the  last  term  diminished  by  the  first  term  and  divided 
by  the  ratio  less  1. 

By  using  the  value  of  I  (Sec.  210), 


s  = 


ar"  1  •  r  —  a     arn  —  a 


r-1  r-1 

Thus,  s  may  be  found  without  first  computing  I. 

WRITTEN    EXERCISES 

1.  Find  the  sum  of  6  terms  of  the  series  ^,  \,  \,  •••• 

2.  Find  the  sum  of  10  terms  of  the  series  \,  —  \,  \,  •••. 

3.  Find  the  sum  of  8  terms  of  the  series  1,  .25,  .0625,  •••. 

4.  Find  the  sum  of  12  terms  of  the  series  27,  —  9,  3,  —  1,—. 

5.  An  air  pump  exhausted  the  air  from  a  cylinder  containing 
1  cu.  ft.  at  the  rate  of  j1^  of  the  remaining  contents  per  stroke. 
What  part  of  a  cubic  foot  of  air  remained  in  the  cylinder  after 
25  strokes  ? 

6.  The  population  of  a  town  increased  from  10,000  to 
14,641  in  5  yr.  If  the  population  by  years  was  in  geometric 
series,  what  was  the  rate  of  increase  per  year  ? 

213.  Collected  Results.  The  three  chief  formulas  of  geo- 
metric series  are : 

1.    l  =  arn~\ 

Ir  —  a 


2.  s  = 

3.  s  = 


r-1 

ar'1  —  a 


r-1 


SERIES 


177 


1st 

8100  (1.05) 

2d 

•$100  (1.05)2 

3rd 

$100  (l.os)3 

4th 

$100  (1.05)* 

WRITTEN    EXERCISES 

1.  $100  is  placed  on  interest  at  5%,  compounded  annually. 

(1)  What  is  the  amount  at  the  end  of  the  first  year  ? 

(2)  What  is  the  principal  for  the  second  year  ? 

(3)  What  is  the  amount  at  the  end  of  the  second  year  ? 

Notice  that  the  amounts  appear  in 
the  right-hand  column  of  the  table. 

(4)  Indicate  similarly  the  amount 
of  $100  at  the  end  of  5  yr. ;  10  yr. ;  n 
yr.  Which  formula  of  geometric  series 
expresses  the  amount  for  n  yr.  ? 

2.  Indicate  the  amount  of  $100  at  6%,  compounded  an- 
nually, at  the  end  of  1  yr.;  2  yr. ;  5  yr. ;  10  yr. ;  n  yr. 

3.  Many  savings  banks  pay  interest  at  the  rate  of  3  %, 
compounded  semiannually. 

Indicate  the  amount  of  $100  under  the  above  conditions 
at  the  end  of  6  mo. ;  1  yr. ;  18  mo. ;  2  yr. ;  10  yr. ;  n  yr. 

Note.  The  numerical  value  of  these  expressions  can  be  computed 
readily  by  logarithms. 

Solve  by  the  use  of  logarithms  : 

4.  What  is  the  amount  of  $  1  at  1  %  compound  interest  for 
8  yr.  ? 

5.  A  man  deposits  $100  in  a  bank  paying  -i</0  interest,  com- 
pounded annually,  on  the  first  day  of  each  year  for  5  yr. 
How  much  will  he  have  on  deposit  at  the  end  of  5  yr.  ? 

6.  Determine  similarly  the  amounts  of : 


Deposit   at  Beginning 
or  Each  Tear 

Kate  of  Interest  Com- 
pounded Annually 

Number  of  Tears 

(1) 

$25 

5 

8 

(2) 

10 

6 

15 

(3) 

43.20 

4 

20 

w 

39.87 

3 

20 

178  ELEMENTARY   ALGEBRA 

MEANS 

214.  Means. .  Terms  standing  between  two  given  terms  of  a 
series  are  called  means. 

215.  Arithmetical  Mean.  If  three  numbers  are  in  arith- 
metical series,  the  middle  one  is  called  the  arithmetical  mean 
between  the  other  two. 

The  arithmetical  mean  between  a  and  b  is  found  thus : 

1.  Let  A  be  the  mean  and  d  the  common  difference. 
Then  the  terms  may  he  written  A  —  d  and  A  +  d. 

2.  Whence,  A  —  d  =  a  and  A  +  d  =  b. 

3.  Adding,  2  A  =  a  +  6  and  A  =  ^±-5  ■ 

216.  The  arithmetical  mean  between  two  numbers  is  one  half 
their  sum. 

217.  Geometric  Mean.  If  three  numbers  are  in  geometric 
series,  the  middle  one  is  called  the  geometric  mean  between  the 
other  two. 

The  geometric  mean  between  a  and  b  is  found  thus : 

1 .  Let  g  be  the  geometric  mean. 

2.  Then  2=1 

a     g 

3.  .-.  g-  =  ab  and  g  —  Vab. 

218.  Tlie  geometric  mean  between  two  numbers  is  the  square 
root  of  their  product.  There  are  really  two  geometric  means,  one 
negative  and  one  positive. 

The  geometric  mean  between  two  numbers  is  the  same  as 
their  mean  proportional. 

ORAL    EXERCISES 

State  the  arithmetical  mean  between : 

l.   8,12.        2.   (5,3.  3.   4,-10.  4.   5a,  13a. 

State  the  geometric  mean,  including  signs,  between : 

5.   8,6.  6.   3,12.  7.   a,  a5.  8.   2  a?,  32  aT. 


SERIES  179 

219.  Any  number  of  means  may  be  found  by  use  of 
formulas  already  given. 

EXAMPLES 

1.  Insert  5  arithmetical  means  between  4  and  12. 

1.  In  this  case  a  —  4,  I  =  12,  and  n  =  7. 

2.  .'.  I  =  a  +  O  —  l)'i  becomes  12  =  4  +  (7  -  l)d. 

12  —  4 

3.  Solving  for  c?.  d  =  — - —  =  1J. 

b 

4.  Adding  ]i  to4,  and  1^  to  that  result,  and  so  on,  the  means 

are  found  to  be  5|,  6|,  8,  9i,  and  lOf. 

2.  Insert  4  geometric  means  between  —  27  and  \. 

1.  In  this  case  a  =  —  27,  /  =  |,  and  n  =  6. 

2.  .-.  Z  =  ar11-1  becomes  \  —  —  27  r5. 

3.  Solving  for  r,    ?'5  = =  —  —  • 

&  9  •  27         35 

Therefore,  r  =—\. 

4.  Multiplying  —  27  by  —  \,  and  multiplying  this  result  by  —  i, 

and  so  on,  the  means  are  found  to  be  9,  —  3,  1,  and  —  \. 

WRITTEN    EXERCISES 

1.  Insert  3  arithmetical  means  between  6  and  26. 

2.  Insert  10  arithmetical  means  between  —  7  and  144. 

3.  Insert  3  geometric  means  between  2  and  32. 

4.  Insert  4  geometric  means  between  —  ^  and  31. 

OTHER    FORMULAS 

220.  Arithmetical  Series.  By  means  of  the  formulas  of  Sec. 
206,  any  two  of  the  five  numbers,  a,  n,  I,  d,  s,  can  be  found  when 
the  other  three  are  given. 

EXAMPLES 

1.    Given  n  =  6,  s  —  18,  I  —  8,  find  a,  d. 

1.   For  these  values,  formulas  (1)  and  (2)  become  : 

8  =  a  +  (<S  -  l)d, 

18=6(«  +  8) 
2 


18q  ELEMENTARY   ALGEBRA 

2.  We  have  thus  two  equations  to  determine  the  two  num- 

bers, a,  d. 
From  the  second  equation,  a  =  —  2. 

3.  Using  this  value  in  the  first  equation,  d  =  2.  • 

2.  Given  a  =  4,  I  =  12,  s  =  56,  find  w  and  d. 

1.  For  these  values  formulas  (1)  and  (2)  become: 

12  =  4  +  (»  -  l)d. 

66  =  *C4+12>. 
2 

2.  .-.  from  the  second  equation,  n  =  7. 

3.  Substituting  in  the  first,  12  =  4  +  6  d, 

4.  therefore  d  =  f . 

3.  Given  n  =  12,  s  =  30,  I  =  10,  find  a,  d. 

1.  Formulas  (1)  and  (3)  become  : 

10  =  a  +  11  <Z. 
30=12(2q+lld)  =  12  a  +  6r,fZ< 

2 

2.  Solving  these  equations  for  a  and  d  : 

a  =  —  5,  and  d  =  |f . 
The  same  results  would  be  found  by  using  formulas  (1)  and  (2),  since 
(3)  is  only  another  form  of  (2). 

221.    The  same  problems  can  also  be  solved  generally;  that 
is,  without  specifying  numerical  values. 

EXAMPLE 

Regarding  n,  s,  d  as  known,  find  a,  I. 

1.  From  (1),  Sec.  206,       I  -  a  =(n  -  Y)d. 

2.  From  (2),  Sec.  200,       a  +  I  =  —  • 

n 

3.  Adding  (1)  and  (2),  2  I  =  (»-  1)  d  +  — • 


Or, 


;  =  ("-!>*  |  s. 


9  e      r(ti \}d      s~\ 

L    Substituting  in  (2)  above,  a  =  "- — J—  +  - 

n       L        2  "J 


_  .s      (w-l)d. 
~n  2 


SERIES 


181 


WRITTEN    EXERCISES 
By  use  of  the  formulas  in  Sec.  206,  find,  the  following 


l. 
2. 

3. 


5. 
6. 

7. 

8. 

9. 

10. 

11. 

12. 

13. 
14. 
15. 
16. 

17. 

18. 

19. 
20. 


Find 

I 
I 

I 
I 


a 

a 

a 
a 


d 
d 
d 
d 


n 

n 
n 


In  Terms  of 


ad  n 
ads 

a  n's 
1 1  n  s 


ad  n 
a  d  I 

an  I 

d  »  I 


dn  I 
d  n  s 

d  I  s 
n  I  s 


an  I 
an  s 
a  I  s 
n  I  s 


ad  I 
ads 

a  I  s 

d  I  s 


Result 


I  =a  +  (w  —  \)d 
I 

I  =  - 


i  [  -  d  ±  V8  ds  +  (2  a  -  d)2} 
2s 


I  = 


n 

s     Q-  \)d 
n  2 


%  n[2a  +  (n-  l)rf] 
1+  a     f  -  a? 


2d 


=  \  n  [2  I-  (n-  l)d] 


a 

a 

a 
a 


I-  («  -  \)d 
s_  (».  -  \)d 
n  2 


=  \  \_d  ±  V(2  I  +  d)2  -  8  ds] 


2  s 


d 
d 
d 
d 


1  —  a 
ii-l 

2  (s  —  an) 
11(11  —  1) 

I2  -  a2 
2  s  —  I  —  a 
2(nl-  s) 
11(11  —  1) 


1  —  a  ,  ■. 
n  = h  1 


d 


ii 


d-2a±  V  (2  a  -  d)2  +  8  ds 


n  =■ 


n  = 


2d 

2  s 

1  +  a 

2  1  +  d  ±  V(2Z  +  c?)2- 

2d 


8ds 


Note,  a,  I,  d,  s  may  have  any  values,  but  n  must  be  a  positive  integer. 
Hence,  when  n  is  one  of  the  unknowns,  not  all  the  solutions  that  satisfy 
the  equations  will  correspond  to  a  possible  arithmetical  series. 


182  ELEMENTARY   ALGEBRA 

222.  Geometric  Series.  By  means  of  the  formulas  of  Sec. 
U13,  any  two  of  the  five  numbers,  a,  n,  I,  r,  s,  can  be  found 
when  the  other  three  are  given. 

EXAMPLES 

1.  Given  s  =  1024,  r  =  2,a  =  2,  find  I. 

1.  For  these  values  formula  (2)  becomes: 

1024  =  2-^?  =  2  (7-1). 
2-1  v         J 

2.  .-.  Z  =  513. 

2.  Given  r  =  3,  n  =  5,  s  =  363,  find  a. 

1.  For  these  values  formulas  (1)  and  (2)  become; 

l=.a-3*. 
31 -a 


363  =  ' 


2 


2.  Eliminating  1,       363=^! — 11. 

3.  Therefore,  363  =  q\242,  and  a  =  3. 

3.    Given  s  =  363,  a  =  3,  r  =  3,  find  n. 

1.  For  these  values  formula  (3)  becomes: 

a  .  sn  _  a 
363  =  5_2 2. 

2 

2.  Therefore,     3"  -  1  =  242,  and  3"  =  243. 

3.  By  factoring  243,    n  is  seen  to  be  6. 

Note.     In  finding  n  it  may  not  be  possible  to  factor  as  in  the  case  of 
243  above.     In  this  case  logarithms  may  be  applied. 

223.    The  same  problems  can  be  solved  generally,  that  is, 
without  specifying  numerical  values. 

EXAMPLE 
Express  I  in  terms  of  a,  n,  and  s. 

1.  From  formula  (2)  r=s-^^,  or,    r»-i=^~^"". 

s-l  is -I)"-1 

Cl  (g ryV>-l 

2.  .-.  substituting  in  (1 )  I  =  — ^ '  '      • 

(.s  —  i)n 

3.  .-.  l(s  -  0n_1  -  a  («  -  a)""1  =  0. 

This  equation  is  of  a  degree  higher  than  2  in  I  when  n  >  3.     But  for 
?i  equal  to  or  less  than  3  it  can  be  solved  by  methods  already  explained. 


SERIES 


183 


WRITTEN    EXERCISES 
By  use  of  the  formulas  in  Sec.  213,  find  the  following: 

Note.     In  Exercises  3,  12,  and  16,  only  the  equation  connecting  the 
unknown  numbers  with  the  given  ones  can  be  found  : 


1 
2. 
3. 
4. 

5. 
6. 

7. 

8. 


10. 

11. 
12. 

13. 
14. 
15. 

16. 


Find 


a 
a 


In  Terms  of 


a  r  n 
ars 

a  n  s 
r  n  s 


a  r  n 
a  rl 

a  n  I 
r  n  I 


r  n  I 

r  ns 

r  Is 

n  1  s 


an! 


a  n  s 


a  1  s 


nls 


Result 


I  =  ar"'1 
_a  +  (r-l).t 
r 

l(S-  O""1  -rt(S  -«)"-!  =  0 

(r  —  l)s?-"-1 
rn  —  1 


s  = 


s  = 


«Q"-1) 

r-1 
rl  —  a 
r-1 

n—li—       n— 1/ — 

VY"  -     Va» 

'-Vl-"-lTa 
lrn  -  I 


yWI    y.H-1 


a  = 


i.n-1 


a  = 


(r  -  1)  s 


r"  —  1 
a  =  rl  —  (r  —  l)s 

a  (s  -  a)""1  -  I  (s  -  I)"-1  =  0 


n-\a 

r=    X 


s        s  —  a     rt 

rn  —  ?•  +  -     -  =  0 
a  a 

s  —  a 


s  -  I 
s 


rn-       —  rn~l  +-  — -  =  0 
s  —  /  s  —  / 


184  ELEMENTARY   ALGEBRA 

SUMMARY 

I.  Definitions. 

1.  A  set  of  numbers  specified  according  to  some  law  is  called 
a  series.     The  numbers  constituting  a  series  are  called  its  terms. 

Sees.  196,  197. 

2.  If  each  term  of  a  series  after  the  first  is  found  by  adding 
a  fixed  number  {common  difference)  to  the  preceding  term,  the 
series  is  called  an  arithmetical  series,  or  arithmetical  progression  ; 
if  each  term  after  the  first  is  found  by  multiplying  the  preceding 
term  by  a  fixed  number  {common  ratio),  the  series  is  called  a  geo- 
metric series,  or  geometric  progression.     Sees.  200,  201,  207,  208. 

3.  If  three  numbers  are  in  arithmetical  (or  geometric)  series, 
the  middle  one  is  called  the  arithmetical  (or  geometric)  mean 
between  the  other  two.  Sees.  440-444. 

II.  Notations.      a  =  first  term. 

d  =  common  difference. 

r  =  common  ratio. 

n  =  number  of   a  term,  or  number  of  terms 

considered. 
I  =  last  (of  n  terms). 
s  =  sum  (of  n  terms). 

III.  Important  Formulas. 


Arithmetical  Series 

Geometric  Series 

I  =  a  -f  (n  —  l)d 

n  (a  +  I) 
S=        2 

o   

2 

Z  =  crr"-1 

Ir  —  a 

a,r"  —  a 

S-   r-1 

REVIEW 

WRITTEN    EXERCISES 

1.  Find  the  47th  multiple  of  7. 

2.  Find  the  sum  of  the  first  12  multiples  of  4. 


SERIES  185 

Find  the  20th  term,  and  the  sum  of  12  terms  of  each  series-. 

4.  8,  11,  14,  17,  .... 

5.  29,  26,  2s,  .... 

6.  a  +  &,  a  —  b,  a  — 3  b,  a  —  5  b. 

Find  the  eighth  term,  and  the  sum  of  8  terms : 

7.  1,  4,16,  -..  10.  1,  -2,  22,  -23,  .... 

8.  3,6,12,.-..  11.  ii,TL   -. 

9.  2,  -4,  8,  -16,  ....  12.  100,  -40,  16,  .... 

Find  the  twelfth  term,  and  the  sum  of  12  terms  : 

13.  2,4,6,  -•.  16.  f,^,^,  .... 

14.  -5,  -3,  -1,  ••-.  17.  4,  -3,  -10,  ••.. 

IK     1      6     5     ...  1Q      1     —1     _1_1      ... 

ID.     -L,    7,  j,    .".  J.O.     2,  3,  g  , 

19.  Find  three  numbers  whose  common  difference  is  1  and 
such  that  the  product  of  the  second  and  third  exceeds  that  of 
the  first  and  second  by  ^. 

20.  The  first  term  of  an  arithmetic  series  is  n2  —  n  —  1,  the 
common  difference  is  2.     Find  the  sum  of  n  terms. 

21.  In  Italy  the  hours  of  the  day  are  numbered  from  1  to 
24.  How  many  strokes  would  a  clock  make  per  day  in  strik- 
ing these  hours  ? 

22.  How  many  strokes  does  a  common  clock  striking  the 
hours  make  in  a  day  ? 

23.  A  man  leases  a  business  block  for  20  years  under  the 
condition  that,  owing  to  estimated  increase  in  the  value  of  the 
property,  the  rental  is  to  be  increased  $50  each  year.  He 
pays  altogether  $19,500.  What  was  the  rental  of  the  first 
year  ?     The  last  ? 

24.  A  railroad  car  starting  from  rest  began  to  run  down 
an  inclined  plane.  It  is  known  that  in  such  motion  the  dis- 
tances passed  over  in  successive  seconds  are  in  arithmetical 
progression.  It  was  observed  that  at  the  end  of  10  sec.  the  car 
had   passed  over  570  ft.  and  at  the  end  of  20  sec.  2340  ft. 

13 


186  ELEMENTARY   ALGEBRA 

from  the  starting  point.  How  far  did  it  run  the  first 
second?  How  far  from  the  starting  point  was  it  at  the 
end  of  15  sec.  ? 

25.  It  is  known  that  if  a  body  falls  freely,  the  spaces  passed 
over  in  successive  seconds  are  in  arithmetical  progression,  and 
that  it  falls  approximately  16  ft.  in  the  first  second  and  48  ft. 
in  the  next  second.  To  determine  the  height  of  a  tower,  a  ball 
was  dropped  from  the  top  and  observed  to  strike  the  ground 
in  4  sec.     Find  the  height  of  the  tower. 

26.  An  employee  receives  a  certain  annual  salary,  and  in 
each  succeeding  year  he  receives  $72  more  than  the  year 
before.  At  the  end  of  the  tenth  year  he  had  received  alto- 
gether $  10,440.  What  was  his  salary  the  first  year  ?  The 
last? 

27.  The  14th  term  of  an  arithmetical  series  is  72,  the  fifth 
term  is  27.     Find  the  common  difference  and  the  first  term. 

28.  A  man  is  credited  $100  annually  on  the  books  of  a  build- 
ing society  as  follows  :  At  the  beginning  of  the  first  year  he 
pays  in  $100  cash.  At  the  beginning  of  the  second  year 
he  is  credited  with  $6  interest  on  the  amount  already  to  his 
credit ;  and  he  is  required  to  pay  $94  in  cash,  making  his 
total  credit  $200.  At  the  beginning  of  the  third  year  he  is 
credited  with  $12  interest,  and  pays  $88  in  cash,  and  so  on. 
How  much  is  his  payment  at  the  beginning  of  the  tenth  year? 
What  is  his  credit  then  ?  How  much  cash  has  he  paid 
altogether  ? 

29.  At  each  stroke  an  air  pump  exhausts  f  of  the  air  in  the 
receiver.  What  part  of  the  original  air  remains  in  the  receiver 
after  the  8th  stroke? 

30.  At  the  close  of  each  business  year,  a  certain  manufac- 
turer deducts  10  <f0  from  the  amount  at  which  his  machinery 
was  valued  at  the  beginning  of  the  year.  If  his  machinery 
cost  $  10,000,  at  what  did  he  value  it  at  the  end  of  the  fourth 
year  ?     . 

31.  By  use  of  logarithms,  find  its  valuation  at  the  end  of  the 
20th  year. 


SERIES  187 


SUPPLEMENTARY   WORK 

Special  Series 

Finite  Series.  So  far  we  have  treated  only  series  with  a  fixed 
number  of  terms.  A  series  which  comes  to  an  end  is  called  a 
finite  series. 

Infinite  Series.  A.  series  whose  law  is  such  that  every  term 
has  a  term  following  it  is  called  an  infinite  series. 

For  example  : 

2,  5,  8,  11,  •••,  239  as  here  written  ends  with  239.  But  the  law  of  the 
series  would  permit  additional  terms  to  be  specified.  In  the  above  ex- 
ample, the  next  following  terms  would  be  242,  245,  etc.  It  is  obvious 
that  however  many  terms  may  have  been  specified,  still  more  can  be 
made  by  adding  3.     The  series  is  thus  unending. 

Similarly,  all  of  the  series  so  far  considered  might  have  been  con- 
tinued by  applying  their  corresponding  laws. 

The  term  "  infinite1'  comes  from  the  Latin  infinities,  and  is  here  used 
with  the  meaning,  unending. 

If  the  coefficients  of  the  binomial  expansion  be  regarded  as  a  series, 

,         (»-  1)     n(n  -  l)Q-2)    n(n  -  1)Q  -  2)(n  -  3) 
'     '^"i-'"  3!  '  '  4! 

they  furnish,  when  n  is  a  positive  integer,  instances  of  series  that  come  to 
an  end  according  to  the  law  of  the  series.  If  n  —  3,  the  series  has  4 
terms,  and  if  n  =  10,  the  series  has  11  terms  ;  for  the  positive  integer  n, 
it  has  Ti  +  1  terms.  This  is  true  because  the  factors  n,  n  —  1,  n  —  2,  and 
so  on,  will  finally  in  the  (n  +  2)nd  term  contain  n  —  n  or  zero.  There- 
fore the  series  has  n  +  1  terms. 

But  if  n  is  a  negative  integer  or  a  fraction,  none  of  the  factors,  n, 
n  —  1,  n  —  2;  and  so  on,  become  zero,  and  the  series  can  always  be  ex- 
tended farther.  That  is,  if  n  is  a  negative  integer  or  any  fraction,  the 
series  is  unending  or  infinite. 


188  ELEMENTARY  ALGEBRA 

Infinite  Geometric  Series.  The  subject  of  infinite  series  is  of 
great  importance,  but  is  far  too  difficult  to  be  taken  up  here. 
We  mention  simply  a  few  properties  of  infinite  geometric 
series  whose  ratio  is  numerically  less  than  1. 

The  following  are  examples  of  such  series  : 
1     4   '2    1    i    i    ... 

9        3      3        3  3 

z>     D5  5'   25'   T251         • 

3.   .5,  .05,  .005,  .0005,  •••. 
4     1_ii—i     i-    _    i 

rx'      A'  2~'   ?'  ?)  T6'  32' 

State  the  ratio  and  the  next  three  terms  of  each  series. 

I.  The  terms  become  numerically  smaller  and  smaller.  Each 
term  is  numerically  smaller  than  the  one  preceding  it,  for  it  is 
a  proper  fraction  of  it. 

II.  Tlie  terms  become  numerically  small  at  will.  That  is, 
however  small  a  number  may  be  selected,  there  are  terms  in 
the  series  smaller  than  it,  and  when  r  is  numerically  less  than 
1,  the  term  arn~l  may  be  made  numerically  small  at  will,  by 
taking  n  sufficiently  large. 

This  seems  obvious  from  the  consideration  of  the  series  given  above  as 
examples.  It  is  not  difficult  to  prolong  these  series  until  their  terms  are  less 
than  T^  say,  or  -j-^g^,  and  from  this  it  seems  plausible  to  think  that  the 
terms  would  become  less  than  one  millionth,  or  one  billionth,  or  any  other 
number,  if  a  sufficient  number  of  terms  are  taken.  As  a  matter  of  fact 
this  is  true,  but  the  proof  is  too  difficult  to  be  given  here. 

III.  We  have  proved  that,  if  sn  denote  the  sum  of  the  first 

n  terms  of  a  geometric  series, 

*    P 
_  a  —  arn 

1  —  r 

This  may  be  written :  sn  = arn~x  [ V 

1  —  r  \1  —  rj 

By  taking  n  sufficiently  large,  the  product  of  arn~x  and  the 

fixed  number can  be  made  as  small  as  desired.     As  more 

1  —  r 

and  more  terms  of  the  series  are  added,  the  sum  differs  less. 


SERIES  189 


and  less  from ;  and  if  sufficiently  many  terms  are  taken, 

1  —  r 

the  sum  comes  as  close  as  we  please  to 

1  —  r 

The  number is  called  the  limit  of  the  sum  of  n  terms, 

1  —  r 

as  n  is  increased  without  bound.     Denoting  this  limit  by  s,  we 

have : 

a 
s  =  - . 


The  number  s  is  not  the  sum  of  all  the  terms  of  the  series,  for  since  the 
terms  of  the  series  never  come  to  an  end,  the  operation  of  adding  them 
cannot  be  completed.  We  cannot  end  an  unending  process.  The  num- 
ber s  is  simply  the  number  to  which  the  sum  of  the  first  n  terms  of  the 
series  approximates  more  and  more  closely  as  n  increases. 

For  example : 

1  4 

When  a  =  4,  and  r  =  -,  then  s  =  — —  =  8. 

2'  1-| 

To  test  this,  we  form  successive  values  of  sn. 

S2  =  6. 

s3  =  7. 
Si  =  7J. 
s5  =  7f . 

s6  =  7|. 

It  appears  that  the  values  of  sn  approximate  more  and  more  closely  to 
8  as  n  is  increased. 

WRITTEN  EXERCISES 
Find  the  limit  of  the  sum  of  the  series  : 
1.    1  +  1+1-+....  3.   5  +  |  +  |  +  ^.... 

2-  W+W+-.  4.  8-I+A-A+-- 

5.  Test  the  results  of  the  preceding  exercises  by  finding 
successive  values  of  sn. 

6.  In  an  infinite  geometric  series  s  =  2  and  r  =  i ;  find  a. 

7.  Find  the  fraction  which  is  the  limit  of  .333333  •••,  or 
.3 +  .03 +  .003+  •-.. 


190  ELEMENTARY   ALGEBRA 

8.   Find  the  limit  of  .23232323  •  •  •  or  .23  +  .0023  +  .000023  +  .... 

,r-  r-~- r— r- — r-r- 1        9.    Triangles  are  drawn  in  a 

Z  ^IfiiiTm^  lilirr^  itiV  Ik  I  rectangle  of  dimensions  indi- 
a  b  d  e  c  cated,  B  being  the  midpoint  ot 
< 12|Ns- ->    Aq  D  that  of  BCf  E  that  of 

DC,  and  so  on.     What  limit  does  the  sum  of  the  areas  of  the 
triangles  approach  as  more  and  more  triangles  are  taken  ? 

ADDITIONAL    EXERCISES 

1.  Find  the  sum  of  16  terms  of  the  series, 

27,  221,  18,  131   .... 

2.  Find  the  sum  of  18  terms  of  the  series, 

3.  The  difference  between  two  numbers  is  48.  The  arith- 
metic mean  exceeds  the  geometric  mean  by  18.  Find  the 
numbers. 

4.  Express  as  a  geometric  series  the  decimal  fraction 

.0373737-.. 
What  is  its  limiting  value? 

5     If  — - —    —    ,  are  in  arithmetical  progression,  show 

b  —  a    2  b    b  —  c 

that  a,  b,  c  are  in  geometric  progression. 

Suggestion.     The  supposition  means  that 
1111 


b  -  a      2b      2f>      b-c 
This  reduces  to  b2  —  ac. 

6.  Find  the  amount  in  n  years  of  P  dollars  at  r  per  cent 
per  annum,  interest  being  compounded  annually. 

7.  During  a  truce,  a  certain  army  A  loses  by  sickness  14 
men  the  first  day,  15  the  second,  16  the  third,  and  so  on ; 
while  the  opposing  army  B  loses  12  men  every  day.  At  the 
end  of  fifty  days  the  armies  are  found  to  be  of  equal  size. 
Find  the  difference  between  the  two  armies  at  the  beginning 
of  the  truce. 


SERIES  191 

8.  A  strip  of  carpet  one  half  inch  thick  and  29f  feet  long 
is  rolled  on  a  roller  four  inches  in  diameter.  Find  how  many 
turns  there  will  be,  remembering  that  each  turn  increases  the 
diameter  by  one  inch,  and  taking  as  the  length  of  a  circum- 
ference --J--  times  the  diameter. 

9.  Insert  between  1  and  21  a  series  of  arithmetic  means 
such  that  the  sum  of  the  last  three  is  48. 

(1  C 

10.  If  -  =  - ,  prove  that  ab  +  cd  is  a  mean  proportional  be- 

(J  Cv 

tween  a2  +  c2  and  b2  +  d2. 

11.  The  sum  of  the  first  ten  terms  of  a  geometric  series 
is  244  times  the  sum  of  the  first  five  terms;  and  the  sum  of 
the  fourth  and  the  sixth  term  is  135.  Find  the  first  term  and 
the  common  ratio. 


CHAPTER   XII 
ZERO:     INTERPRETATION   OF   RESULTS 

ZERO   AND   ITS   PROPERTIES 

224.  Definition  of  Zero.  Zero  may  be  defined  as  the  result 
of  subtracting  a  number  from  itself. 

6  —  6  =  0;  a  —  a  =  0. 

225.  Addition.  By  definition  of  zero,  a  +0=a+b  —  b  =  a, 
since  to  add  b  and  immediately  to  take  it  away  again  leaves 
the  original  number  a. 

226.  Subtraction.  Similarly,  a  —  0  =  a  —  (b  —  b)  =  a  —  b  + 
b  =  a,  since  to  take  away  b,  then  at  once  to  replace  it,  leaves 
the  original  number  a. 

To  add  or  subtract  zero  does  not  alter  the  original  number. 

227.  Multiplication.     By  definition  of  zero, 

0  •  a  —  (b  —  b)a  =  ba  —  ba  =  0. 
That  is,  if  one  factor  is  zero,  the  product  is  zero. 

Multiplication  by  zero  simply  causes  the  multiplicand  to  vanish. 

228.  Division.  We  recall  that  division  is  the  process  of 
finding  a  number  (quotient)  which  when  multiplied  by  a  given 
number  (divisor)  shall  have  a  given  product  (dividend). 

12  -=-  3  or  ±£  simply  proposes  the  problem :  By  what  must 
3  be  multiplied  to  produce  12  ?  The  proof  that  12  -f-  3  =  4  is 
the  fact  that  3  x  4  =  12. 

Likewise  a-=-0,  or  -,  simply  proposes  the  problem,  By  what 

must  zero  be  multiplied  to  produce  a  ? 

Let  x  denote  the  desired  number.     Then  0  •  x  =  a. 

192 


ZERO.     INTERPRETATION   OF   RESULTS  193 

But  we  know  that  zero  times  any  number  is  zero.  If  a  is 
not  zero,  there  is  no  number  x  that  satisfies  the  above  equation. 

That  is,  -  =  no  number,  since  there  is  no  number  whose  prod- 
uct with  0  is  a. 

If  a  is  zero,  every  number  x  satisfies  the  equation.  That  is, 
-  =  any  number,  since  0  times  any  number  =  0. 

Division  by  zero  is  therefore  either  entirely  indefinite  or  im- 
possible.    In  either  case  it  is  not  admissible. 

229  If  we  divide  one  literal  expression  by  another,  there 
is  no  guarantee  that  the  result  is  correct  for  those  values  of 
the  letters  that  make  the  divisor  zero. 


EXAMPLE 

Let 

a  =  b. 

CO 

Multiplying  both  members  by  a, 

a2  =  ab. 

(*) 

Subtracting  b2  from  both  members, 

«2  _  jy2  =  ab- 

■b2. 

(3) 

Factoring, 

(a 

+  b)(a-  b)=b(a- 

-6). 

(4) 

Dividing  both  members  by  a  —  l>, 

a  +  b  =  b. 

{5) 

Substituting  the  value  of  a  from  (i), 

6  +  6  =  6. 

(.6) 

Or, 

2b  =  b. 

(7) 

Dividing  by  b, 

2  =  1. 

(*) 

The  work  is  quite  correct  to  equation  (4)  inclusive.  But  by  dividing 
equation  (4)  by  an  expression  that,  according  to  the  conditions  of  the 
problem,  is  zero,  we  find  as  result  an  incorrect  equation, 

230.  In  all  divisions,  therefore,  we  must  assure  ourselves 
that  the  divisor  is  not  zero.  If  a  literal  divisor  is  used, 
the  result  can  be  depended  upon  only  for  such  values  of  the 
letters  as  do  not  make  the  divisor  zero. 

EXAMPLES 

1.  The  town  B  is  d  miles  distant  from  A;  two  trains  leave 
A  and  B  simultaneously,  going  in  the  same  direction  (that 
from  A  towards  B),  at  the  rate  of  m  mi.  per  hour  (train  from  A) 


194  ELEMENTARY   ALGEBRA 

and  q  mi.  per  hour  (train  from  B).  At  what  distance  from  B 
will  the  two  trains  be  together  ? 

Solving  this  problem  by  the  usual  method,  we  find  as  the  result  —2 

m  —  q 
If  el  =£  0  (read  "  d  is  not  equal  to  0  "'),  and  if  m  =  q,  the  result  assumes 

the  form  2L,  an  indicated  division  by  zero.     This  means  that  the  problem 

is  impossible  under  these  conditions.  This  is  evident  also  from  the  mean- 
ing of  m  and  q  in  the  problem.  If  the  two  trains  go  in  the  same  direction 
at  the  same  rate,  the  one  will  always  remain  d  miles  behind  the  other. 

If,  however,  d  =  0,  and  m  =  q,  the  result  assumes  the  form  - ,  which 

equals  any  number  whatever.  This  also  agrees  with  the  conditions  of  the 
problem.  If  d  is  zero,  B  and  A  are  coincident,  and  the  two  trains  are 
together  at  starting.  If  m=q,  they  both  run  at  the  same  rate,  and  always 
remain  together.     They  are  therefore  together  at  every  distance  from  B. 

2.    We  have  solved  (First  Course,  Sec.  267,  p.  222)  the 

equations 

ax  +  by  =  e, 

ex  +  dy  =f. 

•4-1,4.1            14.             de  —  bf           af—ec 
with  the  result      x  = ■- ;  y  =  -± • 

ad  —  be  ad  —  be 

I.  Let  us  give  the  letters  a,  b,  c,  d,  such  values  that  ad  —  be 
=0;  for  example,  a=2,  6=1,  c=4,  d=2.  And  let  us  give  e  and 
/  such  values  that  de  —  bf  is  not  0;  for  example,  e  =  5,  /=  4. 

Then  the  above  results  become : 

6                12 
x  =  -  :  y  = • 

0'  9  0 

The  indicated  division  by  zero  means  that  the  problem  is  impossible. 
There  is  no  pair  of  values  that  satisfies  both  equations.  This  appears 
readily  also  by  substituting  the  values  of  a  •••/  in  the  given  equations, 
which  then  become  :  2  a;  +  ?/  =  5 

4  x  +  2  y  =  4. 
Dividing  the  second  equation  by  2,  the  system  becomes  : 

2x  +  y  =  6, 

2x  +  y  =  2, 

and  it  is  obvious  that  no  set  of  values  of  x  and  y  can  make  2  x  +  y  equal 
to  5  and  also  equal  to  2. 

The  two  equations  are  called  incompatible  or  contradictory. 


ZERO.     INTERPRETATION   OF   RESULTS 


195 


This  condition  can  be  illustrated  graphically : 


Drawing  the  graphs  of 
2  x  +  y  =  5  and  2  x  +  y  =  2, 
the  two  lines  appear  to  be 
parallel.  That  two  parallel 
straight  lines  do  not  intersect  is 
the  geometric  condition  corre- 
sponding to  the  fact  that  a  system 
of  two  incompatible  equations 
has  no  solution. 


II.  Retaining  the  values 
of  a  •••  d  above,  let  us  give 
e  and  /  such  values  that 
the  numerators  of  the  result 
both  become  zero  ;  for  example,  e  =  5,  /=  10 

The  result  assumes  the  form : 


X" 

•••j 

■■ 

\ 

5 

r 

;•■■■ 

4 

V 

2 
1 

,       \ 

-■CC 

0 

l\         2 

\  3 

0 
x  =  -  ;  y 

0'  J 


0 
0* 


This  indicates  that  x  may  have  any  value  ;  and  also  that  y  may  have 
any  value. 

Substituting  the  values  of  a  ■••  f  in  the  given  equations,  they  become  : 

2x  +  y  =  5 
4x  +  2y  =  10. 

It  appears  that  the  second  equation  is  twice  the  first,  and  hence  equiva- 
lent to  it.  Any  values  of  x  and  y  that  satisfy  the  first,  will  also  satisfy 
the  second. 

We  can  choose  arbitrarily  any  value  for  x  and  then  determine  a  value 
of  y  to  go  with  it  by  means  of  the  first  equation.  For  example,  choosing 
x  =  3,  then  2  •  3  +  y  =  5,  which  gives  y  =  —  1.  These  values  of  x  and  y 
satisfy  both  equations.  Similarly,  any  value  can  be  chosen  for  y,  and  a 
value  of  x  can  be  determined  such  that  the  pair  of  values  satisfies  the  given 
system.  That  is,  any  value  of  x  is  a  root ;  likewise  any  value  of  y  is  a 
root,  in  agreement  with  the  meaning  of  the  form,  -,  assumed  by  the 
result  of  the  general  solution. 

The  two  equations  are  dependent.  Every  solution  of  one  is 
a  solution  of  the  other.  The  conditions  can  be  illustrated 
graphically. 


196 


ELEMENTARY   ALGEBRA 


If  we  undertake  to  make  the 
graphs  of  the  two  equations  as 
given,  we  find  that  they  lead  to 
the  same  straight  line.  The 
two  graphs  are  coincident  ; 
every  point  of  the  straight  line 
is  a  common  point  of  the  two 
graphs.  Any  abscissa  x  is  the 
abscissa  of  a  common  point  of 
the  graphs  ;  any  ordinate  y  is 
the  ordinate  of  a  common  point 
of  the  graphs. 

Note.  The  study  of  expres- 
sions which  may  assume  the 
exceptional  forms  mentioned  above,  especially  those  which  may  assume 
the  form  - ,  is  very  important,  both  from  the  point  of  view  of  later  mathe- 
matics and  the  physical  sciences  ;  but  what  has  been  said  above  will  suffice 
for  the  needs  of  the  present  work. 

231.    We  have  thus  seen  that  systems  of  two  linear  equa- 
tions in  two  unknowns  may  be  classified  as  follows : 

1.  Independent  (the  ordinary  case,  admitting  one  solution). 

2.  Contradictory  (admitting  no  solution). 

3.  Dependent  (admitting  a  boundless  number  of  solutions). 


WRITTEN    EXERCISES 


Construct  the  graphs  of  each  of  the  following  systems  and 
classify  them  according  to  Sec.  231 : 


1.  3x  +  y  =  2, 
x  +  y  =  0. 

2.  2x-y  =  l, 
Ax-2y  =  2. 

3.  s  —  t  =  6, 

4.  x  +2z  =  10, 
x  -f-  3  z  =  11. 


5.  a  =  25, 
y  =  10. 

6.  lQx  +  hy  =  25, 
2x  +  y  =  5. 

7.  7x  +  Uy  =  7, 
x  +  2y  =  2. 

8.  12x-3y  =  8, 
3y—x  =4. 


ZERO.     INTERPRETATION   OF   RESULTS  197 

232.  The  discussions  above  are  instances  of  what  may  be 
called  interpretation  of  results.  That  is,  after  the  conditions 
of  a  problem  have  been  expressed  by  equations,  and  the  equa- 
tions solved,  the  result  must  be  examined  to  s"ee  whether  it  is 
admissible  under  the  conditions  of  the  problem;  the  various 
possible  combinations  of  the  literal  expressions  given  must  be 
discussed ;  and  exceptional  or  noteworthy  sets  of  values  pointed 
out. 

EXAMPLES 

1.  Find  three  consecutive  integers  such  that  their  sum  shall 
be  equal  to  3  times  the  second. 

Solution.     1.    Let  x  =  the  first. 

2.  Then  x  +  1  =  the  second, 

3.  and  x  +  2  =  the  third. 

4.  .-.  x  +  (x  +  1)  +  (x  +  2)  =  3(sc  -f  1),   by    the   condi- 

tions of  the  problem. 

5.  ...  (3  _  3)  (X  +  1)  =  0,  or  0  (:>:  +  1)  =  0. 

Interpretation  of  the  Result.  The  equation  determines  no  particu- 
lar value  of  x  ;  it  exists  for  every  value  of  x.  Consequently,  every  three 
consecutive  integers  must  satisfy  the  given  conditions. 

2.  Find  three  consecutive  integers  whose  sum  is  57,  and  the 
sum  of  the  first  and  third  is  40. 

Solution.     1.   Let  x  =  the  first. 

2.  Then  x  +  1  =  the  second, 

3.  and  x  +  2  =  the  third. 

4.  Then,  x  +  (x  +  1)  +  (x  +  2)  =  57, 

5.  and  x  +  (x  +  2)  =  40,  by  the  given  conditions. 

6.  From  (4),  x  =  18. 

Interpretation  of  the  Result.  This  result  for  x  will  not  satisfy 
equation  (5)  ;  therefore  no  three  consecutive  integers  satisfy  the  condi- 
tions of  the  problem. 

WRITTEN   EXERCISES 

Solve  and  interpret  the  results : 

1.  Fifteen  clerks  receive  together  $150  per  week;  some 
receive  $  8  and  others  $  12  per  week.  How  many  are  there 
receiving  each  salary  ? 


198  ELEMENTARY   ALGEBRA 

2.  A  train  starts  from  New  York  to  Richmond  via  Phila- 
delphia and  Baltimore  at  the  rate  of  30  miles  an  hour,  and 
two  hours  later  another  train  starts  from  Philadelphia  for 
Richmond  at  the  rate  of  20  miles  an  hour.  How  far  beyond 
Baltimore  will  the  first  train  overtake  the  second,  given  that 
the  distance  from  New  York  to  Philadelphia  is  90  miles  and 
from  Philadelphia  to  Baltimore  96  miles  ? 

3.  The  hot-water  faucet  of  a  bath  tub  will  fill  it  in  14 
minutes,  the  cold-water  faucet  in  10  minutes,  and  the  waste 
pipe  will  empty  it  in  4  minutes.  How  long  will  it  take  to  fill 
the  tub  when  both  faucets  and  the  waste  pipe  are  opened  ? 

4.  If  the  freight  on  a  certain  class  of  goods  is  2  cents  per 
ton  per  mile,  together  with  a  fixed  charge  of  5  cents  per  ton 
for  loading,  how  far  can  2000  tons  be  sent  for  $  80  ? 

5.  Find  three  consecutive  integers  whose  sum  equals  the 
product  of  the  first  and  the  last. 

I 


SUPPLEMENT 

GEOMETRIC   PROBLEMS   FOR  ALGEBRAIC  SOLUTION 

The  problems  in  the  following  list  may  be  used  as  supple- 
mentary work  for  pupils  that  have  studied  plane  geometry. 
In  the  body  of  the  Algebra  numerous  problems  have  been  given 
applying  geometric  facts  which  the  pupil  has  learned  in  the 
study  of  mensuration  in  arithmetic.  In  the  following  list 
the  problems  contain  the  application  of  other  relations  and 
theorems  of  geometry,  and  typical  solutions  have  been  inserted 
to  suggest  to  the  pupil  the  method  of  attack. 

LINEAR    EQUATIONS.     ONE    UNKNOWN 

1.  In  a  given  triangle  one  angle  is  twice  another,  and  the 
third  angle  is  24°.     Find  the  unknown  angles. 

Solution.     Let  x  be  one  of  the  unknown  angles, 

then  2  x  is  the  other.  (i) 

Because  the  sum  of  the  angles  of  a  triangle  =  180°, 

x  +  2  x  +  24°  =  180°.  0?) 

Solving  equation  (2), 

x  =  52°  and  2  x  =  104°.  (3) 

The  angles  of  the  triangle  are  52°,  104°,  24°. 

2.  In  a  certain  triangle  one  angle  is  three  times  another, 
and  the  third  angle  is  36°.     Find  the  unknown  angles. 

3.  In  a  given  right-angled  triangle  one  acute  angle  is  f  the 
other.     Find  the  angles. 

4.  In  a  certain  isosceles  triangle  the  angle  opposite  to  the 
base  is  18°.     Find  the  angles  at  the  base. 

5.  The  three  angles  A,  B,  and  C  of  a  given  triangle  are  in 
the  ratio  of  2,  3,  and  5.     Find  the  angles. 

199 


200 


ELEMENTARY   ALGEBRA 


6.    Given  an  angle  A  such  that  a  point  B  situated  on  one 
side  11  in.  from  the  vertex  is  8  in.  distant  from  the  other  side. 
Find  a  poiut  C  on  the  same  side  as  B,  and 
equidistant  from  B  and  the  other  side  of  the 
angle. 

Solution.     Let  A  be   the  given  angle,  then  the 
figure  represents  the  conditions  of  the  problem. 
From  the  similar  triangles  A  CE  and  ABD,  we  have 

AC^AB 
CE     Brf 

11-  a;  _  11 

8  ' 


or, 

From  (1), 
Then, 
and 


x 

88-8x=  11  a\ 
88  =  19  x, 

X  : 


GO 


8  8 

T5' 


(3) 
(4) 


The  distance  CB  is  f  f  in.,  or  4£f . 


7.  Solve  problem  6,  if  the  point  B  is  9  in.  from  A  and  6  in. 
from  side  AD. 

8.  Solve  problem  6,  if  the  point  B  is  a  in.  from  the  vertex 
of  the  angle  and  b  in.  from  the  other  side  of  the  angle. 

9.  Two  points  A  and  B  are  8  in.  apart.  Parallels  are 
drawn  through  A  and  B ;  on  these  parallels  the  points  A'  and 
B'  are  located  on  the  same  side  of  the  straight  line  through  AB 
and  at  distances  6  in.  and  5  in.  from  A  and  B,  respectively. 
Determine  the  point  where  the  line  A'B'  cuts  the  line  AB. 


s^b 


Solution.     Let  C  be  the  desired  point  and  let  BC  —  x. 
Then,  by  similar  triangles, 

BC       AC 


or 


BB>      A  A' ' 

x  _  x  +  8 


(0 
GO 


GEOMETRIC  PROBLEMS  FOR  ALGEBRAIC  SOLUTION     201 


Hence, 


The  point  is  40  in.  from  B. 


6x 

x 


5  x  +  40. 
40. 


(5) 
(-4) 


10.  Solve  the  same  problem  if  yl'and  B'lie  on  opposite  sides 
of  AB. 

11.  Solve  the  same  problem  if  the  distance  AB  is  d,  and 
the  points  A',  B'  lie  on  the  same  side  of  AB,  and  at  distances  a 
and  6  from  A  and  .B  respectively,  with  a>b. 

12.  Solve  the  preceding  problem  if  the  points  A'  and  B'  lie 
on  opposite  sides  of  AB. 

13.  The  three  sides  of  a  triangle  are  11,  9,  12.  A  perpen- 
dicular is  dropped  on  the  side  of  length  11  from  the  opposite 
vertex.  Find  the  lengths  of  the  segments  into  which  the  foot 
of  the  perpendicular  divides  that  side. 


Solution.    Using  the  notations  of  the  figure, 

A>=12>-a* 
and  fc2  =  92-(ll~a;)2. 

From  (i)  and  (2),  122 - x2  =  92-  (11  - x)2. 

Rearranging  (3) ,  122  -  9s  +  ll2  =  22  x. 

Solving  (4),  «  =  ff 

The  other  segment  is  11 
The  segments  are  ff  and  ff . 


x,  or  ff . 


00 

(*) 

(5) 

(4) 

(5) 

(6) 


14.  Solve  the  preceding  problem  if  the  sides  are  4,  7,  9,  and 
the  perpendicular  is  dropped  on  the  side  of  length  7. 

15.  Solve  the  same  problem  if  the  sides  of  the  triangle  are 
a,  b,  c,  and  the  perpendicular  is  dropped  on  the  side  of  length  a. 

14 


202 


ELEMENTARY   ALGEBRA 


16.    The  lower  base  of  a  trapezoid  is  12,  the  upper  base  is  10, 
and  the  altitude  is  4.     Determine  the  altitude  of  the  triangle 
formed  by  the  upper  base  and  the  prolonga- 
tion of  the  two  non-parallel  sides  until  they 
meet. 


E 
r* 

/» 


10 


Df 


l  i 
l  l 

i  i 
i  \ 
i    • 


or 


Av_ 


or 


JB 


12 


Solution.     Using  the  notations  of  the  figure, 

EF^BC 
EG      A& 

x      _10 

z  +  4~12' 

Hence,  12  x  =  10  x  +  40, 

x  =  20. 
The  altitude  is  20. 


(-0 

(3) 

00 


17.    Solve  the  same  problem  if  the  lower  base  is  a,  the  upper 
base  b,  and  the  altitude  h. 


LINEAR   EQUATIONS.     TWO   UNKNOWNS 

18.  A  rectangle  5  in.  longer  than  it  is  wide  is  inscribed  in  a 
triangle  of  base  12  in.  and  altitude  9  in.,  the  longer  side  resting 
on  the  base   of   the  triangle.      Find   the    dimensions   of   the 


rectangle. 
Solution. 
Then, 


Let  x  denote  the  longer  side  and  y  the  shorter. 
x  —  y  =  5. 


W 


E< 

/' 

ft 

L 

F               \ 

tP 

/ 

\ 

i 

1 

V 

V 

X 

G 

\ 

m                                                  j 

or 


12 

In  the  similar  triangles  ABC  and  AED, 

AF=ED 
AG      BC' 

0—  y  _  x 
9         12* 


(3) 


GEOMETRIC  PROBLEMS  FOR  ALGEBRAIC  SOLUTION     203 


From  (3), 

From  (1)  and  (£), 


108  -  12  y  =  9  x. 
108 -12  ?/ =  9Q/ +  5), 
or  21  y  =  63. 

y  =  s. 

From  (7)  and  (i),  x  =  S. 

The  dimensions  of  the  rectangle  are  3  in.  and  8  in. 


(.4) 
(5) 
(6) 
(?) 
(*) 


19.  Solve  the  preceding  problem  if  the  shorter  side  rests  on 
the  base  of  length  12  in. 

20.  Solve  Problem  18  if  the  difference  of  the  sides  is  d,  the 
length  of  the  base  of  the  triangle  is  a,  the  altitude  is  h,  and 
the  longer  side  of  the  rectangle  rests  on  the  base  of  the  triangle. 

21.  Solve  the  preceding  problem  if  the  shorter  side  of  the 
rectangle  rests  on  the  given  base  of  the  triangle. 

22.  A  rectangle  similar  to  a  rect- 
angle whose  sides  are  5  and  8  is 
inscribed  in  a  triangle  of  base  32 
and  altitude  20.  The  longer  side 
of  the  rectangle  rests  on  the  given 
base  of  the  triangle.  Find  the  di- 
mensions of  the  rectangle. 

Solution.     Let  x  and  y  denote  the  sides  of  the  inscribed  rectangle. 
Then  from  the  similar  triangles  EDC  and  ABC, 

DE  _  CF 

AB~  CG' 

x  _20-y 

32 _     20 
From  the  similarity  of  the  rectangles, 

x     8 


or 


a) 


(3) 
(4) 

(6) 
(7) 

(•'0 

23.    Solve  the  preceding  problem  if  the  shorter  side  of  the 
rectangle  rests  on  the  given  side  of  the  triangle. 


y     5 

From  (5), 

x  =  %y. 

From  (;?), 

20  x  =  640  -  32  y. 

From  (4)  and  (;7), 

32  y  =  640-32?/, 

64  y  =  640. 

y  =  io. 

From  (S)  and  (A), 

x  =  16. 

-04  ELEMENTARY   ALGEBRA 

24.  Solve  the  same  problem  if  the  given  side  of  the  triangle 
is  of  length  a,  and  the  altitude  on  it  is  of  length  h,  and  the 
given  rectangle  has  dimensions  1  and  m,  provided  the  inscribed 
rectangle  has  its  side  corresponding  to  the  side  I  of  the  given 
rectangle  resting  on  the  given  base  of  the  triangle. 

25.  The  bisector  of  an  angle  of  a  given  triangle  divides  the 
side  opposite  to  the  angle  into  two  segments  of  lengths  4  in. 
and  7  in.  The  difference  between  the  other  two  sides  of  the 
triangle  is  5  in.     Find  the  perimeter  of  the  triangle. 


Solution.  Let  ABC  be  the  given  triangle,  AD  the  bisector  of  angle 
A,  and  x  and  y  the  required  sides. 

Then,  x  —  y  =  5,  given  in  the  problem,  •  (l) 

x      7 
and  -  =  -,  by  geometry  the  bisector  divides  the  opposite  side  into  segments 

proportional  to  the  adjacent  sides.  rg\ 

4x  =  7y,  from  (,?).  (5) 

7  y  -  4  y  =  20,  from  (3)  and  4  times  (2)".  (4) 

Then,  y  =  6§,  solving  (4).  (5) 

Then,  .,•  =  llf,  from  (5)  and  (/).  (6) 

The  perimeter  is  11  in.  +  6|  in.  +  llf  in.  =  29i  in. 

26.  Solve  the  preceding  problem,  if  the  segments  of  the  base 
are  I  and  m  and  the  difference  between  the  sides  is  d. 

27.  The  sides  of  a  triangle  are  8  ft.,  12  ft.,  and  15  ft,  and 
the  angle  between  the  sides  8  and  12  is  bisected  by  a  line  cut- 
ting the  side  15.  What  is  the  length  of  each  segment  of  the 
line  15? 


GEOMETRIC  PROBLEMS  FOR  ALGEBRAIC  SOLUTION     205 


LINEAR   EQUATIONS.     THREE    UNKNOWNS 

28.    The  points  A,  B,  and  O  are  situated  so  that  AB  =  8  in., 
BC=6  in.,    .10=5  in.      Find  the 
radii    of    three  circles    having   the 
three  points    as    centers    and   each 
tangent  to  the  other  two  externally. 

Solution.     Let  .r,  ?/,  z,  be  the  radii  of 
the  three  circles  as  indicated  in  the  figure. 


Then, 


x  +  y  =  8, 
x  +  z  —  5, 
V  +  z  =  6. 


Adding  (i),  (J),  and  (<?),  and  dividing  the  result  by  2, 


x  +  y  +  z  -  ■*£. 


From  (5)  and  (4), 
From  (,J)  and  (4), 
From  (i)  and  (^), 
The  radii  are  3£  in.,  41  in.,  and  H  in. 


y  =  h 


z  = 


(4) 
(-5) 

(7) 


29.    Solve    the    same    problem   if    AB  =  12,    50=16,    and 
.10=20. 


30.    Solve  Problem  28,  if  BC=  %  AC=  b 
a 


AB  =  c. 


31.  A  triangle  .ISO  is  circumscribed 
about  a  circle  and  its  sides  are  tangent  at 
points  E,  D,  F.  The  sides  of  the  triangle 
are  a =  8  in.,  b  =  15  in.,  and  c  =  12  in.  Find 
the  segments  into  which  points  E,  D,  F 
divide  the  sides. 

Solution.  The  tangents  from  an  external  point 
are  equal,  hence  in  the  figure, 

x  +  y  =  8,  (1) 

x  +  z  =  15,  (..') 

II  +  Z  =  12.  (J) 


Solving  these  equations,  the  segments  are  x 


y 


z  = 


206 


ELEMENTARY   ALGEBRA 


32.  Solve  the  same  problem,  if  a  =  24  ft.,  b  =  10  ft.,  and 
c  =  24  ft. 

33.  A  man  owned  a  triangular,  unfenced  field  of  sides  600  yd., 

700  yd.,  and  800  yd.  He  sold  a  tri- 
angular piece  cut  off  by  a  straight 
line  parallel  to  the  side  of  length 
800  yd.  and  found  that  1800  running 
yards  of  fence  would  be  required  to 
inclose   what    remained.      Find    the 

\)B  lengths  of  the  sides  of  the  portion 
sold. 

Solution.     Using  the  notations  of  the  figure, 

AB  +  BC  +  AC  +  2  DE  =  AB  +  BE  +  EC  +  CD  +  DA  +  2  DE 

=  (AB  +  BE  +  ED  +  DA)  +  (EC  +  CD  +  DE). 

2100  +  2x  =  1800  +(x  +  y  +  z). 

CD  =  DE 
CA     AB' 


Hence, 
But 


and 


or 


and 


Hence, 

From  (7)  and  (8), 

From  (9)  and  (2), 
Hence, 


CE . 
CB 

DE 
~ AB' 

y 

700 

% 
~800' 

z 
600 

_     X 

~800" 

y 

7x 

~  8  ' 

z 

_'6x 
'  4  ' 

21* 

x  +  y  +  z  = 


8 


2100+2  2  =  1800  + 


21  x 


ox 


8 


=  300. 


From  (12)  and  (7), 
From  (12)  and  (8), 

The   lengths   of  the   sides  are 
CD  =  420  yd. 


x  =  480. 
y  =  420. 
z  =  360. 


00 

(*) 

(3) 

(4) 
(5) 
(6) 
(7) 
(*) 

(IS) 

(U) 


Z>£'  =  480yd.,    #(7  =  360  yd.,    and 


GEOMETRIC  PROBLEMS  FOR  ALGEBRAIC  SOLUTION     207 


34.  Solve  Problem  33  if  the  division  line  runs  parallel  to  the 
side  of  length  700  yd. 

35.  Solve  Problem  33  if  the  division  line  runs  parallel  to  the 
side  of  length  600  yd. 

36.  Solve  Problem  33  if  the  lengths  of  the  sides  are  a,  b,  c, 
with  the  division  line  parallel  to  a,  and  the  length  of  fence 
required  is  2 p. 

QUADRATIC   EQUATIONS 

37.  The  sides  of  a  triangle  are  AC  =  T,  BC  =  9,  and  AB  =  10. 
Calculate  the  length  of  the  altitude 
on  the  side  10,  and  of  the  two  seg- 
ments into  which  the  altitude  divides 
that  side. 

Solution.     Using  the  notations  of  the 
figure, 

V) 
0*) 

(3) 

(4) 
(5) 

(«) 

(7) 


h2  =  72  -  x2, 

and 

A2  =  92  _  (10  _  jc)2. 

Then, 

72_x2  =  92-(10-x)2, 

or 

72  =  92-  100  +  20  x 

Then, 

68  =  20x, 

and 

x  =  V  ; 

also, 

10  -  x  =  A*. 

By  (i), 

/i2  =  72_(l_r-,2. 

Then, 


V7'2  •  52  -  172 


_  V(35  +  17)  (35-  17) 
5 

Vo2  •  18 


V'2-20-2-9       6a/26 


5 


5 
6  a/26 


17  S3 

The  segments  are  —  and  -  -  and  the  altitude  is 

5  6  5 


(9) 


208  ELEMENTARY   ALGEBRA 

38.  Calculate  similarly  the  length  of  the  altitude  on  the 
side  of  length  7,  and  of  the  segments  into  which  the  altitude 
divides  that  side. 

39.  Calculate  similarly  the  length  of  the  altitude  on  the 
side  of  length  9,  and  of  the  segments  into  which  the  altitude 
divides  that  side. 

40.  If  the  lengths  of  the  sides  of  a  triangle  are  a,  b,  c,  calcu- 
late the  lengths  of  the  segments  into  which  each  side  is  divided 
by  the  altitude  on  that  side. 

41.  In  a  circle  of  radius  10,  a 
chord  is  drawn  at  distance  6  from 
the  center.  Find  the  radius  of  a 
circle  that  is  tangent  to  the  circle, 
to  the  chord,  and  to  a  diameter  per- 
pendicular to  it. 

Solution.     Using  the  notations  of  the  figure  : 

EC'2  =  EF'2  +  FC'2.  (i) 

(10  -  x)2  =  x2  +  (6  +  x)*.  {2) 

100  -  20  x  +  x2  =  x2  +  36  -)-  12x  +  x2.  (3) 

x2  +  32  x  -  64  =  0.  (4) 

x  =  _32W322+4Tg  (5) 


_      32±16V4!TT 

-  2 {6) 

=  — 16±8V5.  (7) 

The  negative  value  of  x  being  inadmissible  under  the  geometric  condi- 
tions, we  have : 

x=-16  +  8Vo.  (5) 

42.  Solve  the  same  problem,  if  the  radius  of  the  given  circle 
is  12  ft.  and  the  chord  is  4  ft.  from  the  center. 

43.  Solve  Problem  41,  if  the  radius  of  the  given  circle  is 
/•  and  the  chord  is  at  d  distance  from  the  center. 


GEOMETRIC  PROBLEMS  FOR  ALGEBRAIC  SOLUTION     209 


44.  A  point  P  is  selected  on  a  diameter  of  a  circle  of  radius 
6,  at  the  distance  1  from  the 
center.  At  P  a  perpendicular 
is  erected  to  the  diameter  in 
question,  and  a  tangent  is 
drawn  to  the  circle  such  that 
the  point  of  contact  of  the 
tangent  bisects  the  segment  of 
the  tangent  lying  between  the 
perpendicular  and  the  diame- 
ter produced.  Find  the  dis- 
tance from  the  center  to  the 
point  where  the  tangent  cuts 
the  diameter  produced. 


Solution.    Let 


CE  =  x. 


Then  in  the  right  triangle  CDE, 

DE2  =  x2  -  62. 

From  the  similar  triangles  DC'E  and  DGE, 

DE  _  GE 
CE      DE' 

or  DE2  =CE-GE 

=  x  ■  GE. 
PE 


Since  D  bisects  FE, 


GE  = 


2 
x  —  1 


or 


From  (2),  (5),  and  (7),  x(x  ~  l)  =  x2  -  62, 

x2  -  x  =  2  x2  -  72. 
.-.  x2  +  x  -  72  =  0. 


x  =  — 


1  ±  Vl  +  288 


:-0,   8. 


C-0 

(*) 

(3) 

(4) 
(5) 

(6) 

(7) 

(^) 

(9) 
(10) 

{11) 
(12) 


The  negative  root  also  indicates  a  solution.  It  means  that  a  second 
tangent  satisfying  the  required  conditions  cuts  the  diameter  produced  on 
the  opposite  side  from  F,  at  the  distance  9. 


210 


ELEMENTARY   ALGEBRA 


45.  Solve  the  same  problem  if  the  radius  is  12  and  the  point 
lies  at  the  distance  2  from  the  center. 

46.  Solve  the  same  problem  if  the  radius  is  15,  and  the  dis- 
tance of  P  from  the  center  is  35. 

47.  Solve  the  same  problem  if  the  radius  is  r  and  the  dis- 
tance from  the  center  is  d. 

48.  The  tangent  to  a  circle  is  a  mean  proportional  between 
the  segments  of  the  secant  from  the  same  point.  Find  the 
length  of  the  tangent,  if  the  segments  of  the  secant  are  4  ft. 
and  9  ft. 

49.  Two  chords  AB  and  CD  intersect  at  0  within  the  circle. 
The  product  of  OA  and  OB  equals  the  product  of  OC  and  OD. 
Given  OA  =  4,  OB  =  8,  and  CD  =  12,  find  OC  and  OD. 

50.  The  owner  of  a  triangular  lot  whose  sides  are  70,  88, 
and  140  rd.  in  length  wishes  to  divide  it  by  a  straight  fence 

into  two  parts  that  shall  be 
equal  in  area  and  also  have  the 
same  perimeter.  If  the  fence 
connects  the  sides  of  length  70 
and  140  rd.,  how  must  it  be 
placed  ? 

Solution.     Let  DE  be  the  desired 
position  of  the  fence. 


Then, 


AABC  =  2ADBE. 


Since  the  triangles  have  one  angle  in  common, 
A  ABC      70-140 


or,  by  (1), 


£\DBE 

2: 


xy 
70-140 


or 


xy 
xy  =  35-  140. 


By  the  conditions  of  the  problem, 

BD  +  BE  +  DE  =  DA  +  AC  +  CE  +  ED. 


GO 

(J) 

(5) 
(4) 

(5) 


Subtracting  DE  from  both  members  and  replacing  the  other  lines  by 
their  values, 


GEOMETRIC  PROBLEMS  EOR  ALGEBRAIC  SOLUTION     211 


x  +  y  =  70  -  y  +  1-40  -  x  +  88, 

(6) 

2(x  +  y)  =  298, 

(?) 

x  -f-  y  -  149. 

(*) 

From  (4),                              4  xy  =  4  •  35  •  140 

=  I402. 

(9) 

Squaring  (8)  and  subtracting  (9)  from  the  result, 

(x  -  y)2  =  1492  - 1402 

(10) 

=  (149+  140)(149-140) 

(U) 

=  289  •  9. 

(12) 

Then,                                  x  —  y  =  ±  17  ■  3 

=  ±51. 

(13) 

From  (8)  and  (13),                  x  =  100,  49. 

(U) 

y  =  49,  100. 

(15) 

These  values  satisfy  the  algebraic  equations,  but  in  the  concrete  prob- 
lem y  —  100  is  inadmissible,  since  y  lies  on  the  side  of  length  70.  Hence, 
in  the  concrete  problem,  the  result  is  x  —  100,  y  =  49. 

51.  Solve  the  same  problem  if  the  fence  connects  the  sides 
of  length  70  and  88. 

52.  Solve  the  same  problem  if  the  fence  connects  the  sides 
of  length  88  and  140. 

53.  Solve  the  same  problem  if  the  sides  are  of  length  a.  b, 
c,  and  the  fence  connects  the  sides  of  lengths  a 
and  c. 

54.  Find  the  sides  of  a  right-angled  triangle, 
given  its  area  25,  and  its  perimeter  30. 

Solution.     Let  x,  y,  z  denote  the  sides  of  the  triangle, 
z  being  the  hypotenuse. 


Then, 


and 


x2  +  y2  =  z2, 
x  +  y  -f  z  =  30, 

^  =  25. 
2 


Multiplying  both  members  of  (3)  by  4, 

2xy  =  100. 

Adding  (4)  and  (1),         x2  +  2  xy  +  y2  =  z2  +  100, 
or  (x  +  y)2  =  z2  +  100. 

From  (2),  x  +  y  =  30  —  z, 

or  t»  +  y)2  =  C30-»)2. 


(4) 

(5) 

(6) 
(7) 

(8) 


212  ELEMENTARY   ALGEBRA 

From  (8)  and  (6),  (30-  z)2  =  z-  +  100.  (3) 

(i0) 

From  (7),  x  +  y  =  5^,  (i$) 


or 


(30- 

-*>» 

=  :-  +  100. 

900 

-6O2 

+  Z2 

=  s2  +  100. 

6O2 

=  800. 

z 

—     3    • 

: 

'■  +  y 

—  5_a 

3   ' 

X2 

+  2  ay  +  y2  : 

—  ipOO 

—  9      • 

Multiplying  (4)  by  2  and  subtracting  the  result  from  (14), 

X2-2.nj  +  y*  =  1<1Q,  (25) 


Adding  (13)  and  (16)  and  dividing  the  result  by  2, 

2o±5V7 


.'■  = 


3 

Subtracting  (16)  from  (13)  and  dividing  the  result  by  2, 


(17) 


y  =  ^^-  m 


The  sides  are  25  +  5^,  .25-5V7     and   40 
3  3         '  3 

55-.    Solve  the  same  problem  if  the  area  of  the  triangle  is  64. 
and  the  perimeter  48. 


INDEX 


The  numbers  in  the  Index  refer  to  pages  in  the  Book 


Absolute  Terms,  26. 
Addition,  1. 

Commutative  Law  of,  2. 

Associative  Law  of,  •>. 

of  Fractions,  16. 

of  Radical  Expressions,  44. 

Method  of,  31. 
Algebraic  Expressions,  8. 
Alternation,  152. 
Associative  Law,  3,  12. 

Base,  79. 

Binomial  Equations,  116. 

Characteristic,  79,  80. 
Common  Difference,  171. 
Common  Ratio,  174. 
Commutative  Law,  2,  11,  12. 
Completing  the  Square,  36. 
Complex  Numbers,  91. 
Composition,  153. 
and  Division,  153. 

Degree  of  Equations,  26. 
Difference,  4. 
Discriminant,  109. 
Distributive  Law,  12. 
Division,  14. 

Law  of  Exponents  in.  54. 

of  Fractions,  19. 

of  Imaginaries,  93. 


Equations,  25. 
Identical,  25. 
Conditional,  25. 
Solving,  26,  27,  31,  32,  36. 
Roots  of,  25. 
of  One  Unknown,  25. 
of  Two  Unknowns,  31, 132. 
Systems  of,  31. 
Simultaneous,  31,  132. 
Equivalent,  26. 
Degree  of,  26. 
Linear,  26. 
Quadratic,  26,  36. 
Higher,  26,  141. 
with  Three  or  More  Unknowns, 

no 
OO. 

Radical,  48. 
Binomial,  116. 
Incompatible,  194. 
Contradictory,  194. 
Dependent,  195. 
Exponents,  53. 
Laws  of,  53. 
Fractional,  56,  65. 
Negative,  63,  65. 
Zero,  65. 

Factoring,  20,  109. 

Applied  to  Equations,  125. 
Factors,  11. 
Factor  Theorem,  125. 


213 


214 


ELEMENTARY   ALGEBRA 


Forms,  26,  36. 

Linear,  26. 

Quadratic,  36. 
Formulas,  32, 102,  175. 
Fourth  Proportional,  148. 
Fractions,  15. 

Numerator    and    Denominator 
of,  15. 

Terms  of,  15. 

Reduction  of,  16.  , 

Processes  with,  16-19. 

Complex,  19. 

Graphs,  1,  2,  3,  5,  11,  12,  15,  32, 
97,  111,  129,  145,  163,  195. 

Imaginary  Numbers,  90. 

Processes  with,  91. 

Powers  of,  94. 

as  Roots  of  Equations,  94. 
Interpretation  of  Results,  197. 
Irrational  Numbers,  42. 

Logarithms,  74. 
Linear  Equations,  26. 
Linear  Forms,  26. 

Mantissa,  79. 

Mean  Proportional,  148,  153. 

Means,  178. 

Arithmetical,  178. 

Geometric,  178. 
Multiplication,  11. 

Commutative  Law  of,  11. 

Law  of  Exponents  in,  53. 

of  Relative  Numbers,  13. 

Distributive  Law  of,  12. 

of  Fractions,  17. 

of  Radicals,  45. 

of  Imaginaries,  92. 

Negative  numbers,  5. 


Notation  of  Positive  and  Nega- 
tive Numbers,  6. 

of  Polynomials,  125. 
Numbers,  5,  15. 

Relative,  5,  7,  8. 

Positive,  5. 

Negative,  5. 

Rational,  42. 

Irrational,  42. 

Real,  90. 

Imaginary,  90. 

Complex,  91. 

Order  of  Operations,  3. 

Parentheses,  9. 

Notation  of,  125. 
Positive  Numbers,  6. 
Powers,  5. 

Law  of  Exponents  for,  54. 
Product,  11. 
Progression,  171. 

Arithmetical,  171. 

Geometric,  174. 
Proportion,  148. 
Proportional,  148. 

Fourth,  148. 

Third,  148. 

Mean,  148. 

Quadratic  Equatious,  36,  100. 
Complete,  36. 
Incomplete,  36. 
General  Form  of,  100. 
General  Solution  of,  102. 
Literal,  103. 
Simultaneous,  132. 

Radical  Expressions,  42. 
Properties  of,  42. 
Addition  of,  44. 
Subtraction  of,  44. 


INDEX 


215 


Multiplication  of,  45. 

Square  root  of,  51. 

In  Equations,  48,  123. 
Radicals,  42. 
Ratio,  148. 

Common,  174. 
Rational  Numbers,  42. 
Rationalizing  Factors,  47. 

the  Denominator,  47. 
Real  Numbers,  90. 
Reciprocal,  18. 
Relative  Numbers,  5,  7,  8. 
Roots  of  Equations,  25. 

Relation  of,  to  Coefficients,  107. 

Series,  170. 

Arithmetical,  171,  179. 

Sum  of,  172,  17-">. 

Geometric,  174,  182. 

Finite,  187. 

Infinite,  187. 
Signs,  6. 

of  Operation,  6. 

of  Character,  6. 
Square  Roots,  45,  46. 


Substitution,  31. 

Method  of,  31. 
Subtraction,  4. 

Method  of,  31. 
Subtraction,  of  Fractions,  16. 

of  Radicals,  44. 

of  Imaginaries,  92. 
Surds,  42. 

Terms,  170. 

Absolute,  26. 

in  Series,  170,  172,  175. 
Testing,  27,  108. 
Third  Proportionaly-448.  ■  — 

Unknowns,  26. 

Value,  5. 

Absolute,  6. 
Variables,  160. 
Variation,  160. 

Direct,  160. 

Relation  of,  to  Proportion,  160. 

Inverse,  161. 

Zero,  192. 


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f 


22  1941 


Ml 


*WF  S 


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MAR    19  WO 


SEP*'?1941 


OCT    14  1946 


■j  ■ 


FEB  14  1942 


APR  m  m2 


n« 


NOV  13  1944 


^EC 


mov    301944 


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